Haskell, 60 bytes

snd.([]%)
r%(h:t)=max(r%t)$(r++[h])%drop 1t
r%_=(sum r<$r,r)

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The helper function % recursively branches on choosing whether the include the first element and drop the second, or to skip the first element. It takes the maximum of all outcomes, which are tuples whose first element is the sum, and whose second element is the corresponding list which is extracted for the output.

To handle the rule that the empty list is disallowed even if it would have the smallest trick, we do a cute trick of writing sum r<$r rather than sum r.This makes a list whose elements all are sum r and whose length is that of r. That way, when we choose the maximum, we prioritize any list over an empty r, but otherwise comparisons depend on the first element which is sum r .

R, 136 125 bytes

function(l,G=unlist(Map(combn,list(y<-seq(a=l)),y,c(function(x)'if'(all(diff(x)>1),l[x],-Inf)),F),F))G[which.max(Map(sum,G))]

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-6 bytes thanks to digEmAll, who incidentally also outgolfed me.

Returns the shortest subsequence as a list, breaking ties on lexicographically first by indices.

Brute-force generates all index subsequences, then Filters for those that are non-adjacent, i.e., where all(diff(x)>1). Then subsets [ into l using these indices, selecting [[ the first one where the sum is the max (which.max).

I'm pretty sure this is the first R answer I've ever written that uses Filter! sad, Filter is ungolfy, no wonder I've never used it...

05AB1E, 14 bytes

Saved 1 byte thanks to Kevin Cruijssen

ā<æʒĆ¥≠W}èΣO}θ

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or as a Test Suite

Explanation

ā< # push [0 ... len(input)-1]
 æ # compute powerset
 ʒ } # filter, keep lists where:
 ≠W # no element is 1 in the
 ¥ # deltas
 Ć # of the list with the head appended
 è # index into the input with each
 ΣO} # sort by sum
 θ # take the last element

Brachylog (v2), 14 bytes

~ba~c∋₁ᵐᶠ+ᵒt

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Function submission; input from the left, output from the right, as usual. Very slow; a five-element list is probably the maximum for testing on TIO.

~ba~c∋₁ᵐᶠ+ᵒt
 ~b Prepend an arbitrary element to the input
 a Take a prefix or suffix of the resulting list
 ~c Ordered partition into contiguous sublists
 ∋₁ Take the second element
 ᵐ of each sublist
 ᶠ Find all possible ways to do this
 +ᵒ Sort by sum
 t Take the greatest

The results we get from prefixes aren't incorrect, but also aren't interesting; all possible results are generated via taking a suffix (which is possibly the list itself, but cannot be empty), but "suffix" is more verbose in Brachylog than "prefix or suffix", so I went with the version that's terser (and less efficient but still correct). The basic idea is that for each element we want in the output list, the partition into contiguous sublists needs to place that element and the element before into the same sublist (because the element is the second element of the sublist), so two consecutive elements can't appear in the result. On the other hand, it's fairly clear that any list without two consecutive elements can appear in the result. So once we have all possible candidate lists, we can just take the sums of all of them and see which one is largest.

Husk, 11 bytes

►Σ†!¹mü≈tṖŀ

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Explanation

►Σ†!¹mü≈tṖŀ Implicit input, say L=[4,5,3,4].
 ŀ Indices: [1,2,3,4]
 Ṗ Powerset: [[],[1],[2],[1,2],..,[1,2,3,4]]
 t Tail (remove the empty list): [[1],[2],[1,2],..,[1,2,3,4]]
 m For each,
 ü de-duplicate by
 ≈ differing by at most 1.
 For example, [1,2,4] becomes [1,4].
 † Deep map
 !¹ indexing into L: [[4],[5],[4],..,[5,4],[4,3]]
► Maximum by
 Σ sum: [5,4]

Jelly, 16 14 bytes

JŒPḊf’$ÐḟịµSÞṪ

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Thanks to @EriktheOutgolfer for saving 2 bytes!

JavaScript (ES6),  138 132 130 129  126 bytes

Outputs key-value pairs.

a=>a.reduce((a,x,i)=>[...a,...a.map(y=>[[x,i],...y])],[[]]).map(m=a=>a.some(s=p=([v,i])=>p-(s=~~s+v,p=i)<2)|s<m||(r=a,m=s))&&r

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Step 1

We first compute the powerset of the input with $[value, index]$ pairs.

a.reduce((a, x, i) => // for each value x at position i:
 [ // update a[] to a new array consisting of:
 ...a, // all previous entries
 ...a.map(y => // for each value y in a[]:
 [[x, i], ...y] // append [x, i], followed by all original entries
 ) // end of map()
 ], // end of new array
 [[]] // start with a = [[]]
) // end of reduce()

Step 2

We then look for the maximum sum $m$ among these sets, discarding sets with at least two adjacent elements. The best set is stored in $r$.

.map(m = // initialize m to a non-numeric value
 a => // for each entry a[] in the powerset:
 a.some(s = p = // initialize s and p to non numeric values
 ([v, i]) => // for each value v and each index i in a[]:
 p - ( // compute p - i
 s = ~~s + v, // add v to s
 p = i // update p to i
 ) < 2 // if p - i is less than 2, yield true
 ) | // end of some()
 s < m || // unless some() was truthy or s is less than m,
 (r = a, m = s) // save a[] in r[] and update m to s
) && r // end of map(); return r[]

T-SQL, 122 119 bytes

Input is a table variable.

This query picks all elements from the table variable, combining these with all non-adjacent elements with higher position values and show the text generated for the highest sum of these values.

WITH C(y,j,v)as(SELECT*,x*1FROM @
UNION ALL
SELECT y+','+x,i,v+x
FROM @ JOIN C ON j+1<i)SELECT
TOP 1y FROM C ORDER BY-v

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Haskell, 81 80 bytes

snd.maximum.map((,)=<<sum).tail.f
f(a:b:c)=f(b:c)++map(a:)(f c)
f a=[]:map(:[])a

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f builds all valid subsequences by either skipping the next element (f(b:c)) or using it and skipping the next (map(a:)(f c)) and recursively work on the rest. For the result, build all subsequences (f), drop the empty subsequence (which occurs first in the list: tail), make pairs (<sum>,<subsequence>) (map((,)=<<sum)), find the maximum (pairs are compared in lexicographical order) -> maximum) and drop the sum (snd).

Edit: -1 byte thanks to @Lynn.

Wolfram Language (Mathematica), 144 bytes

If[Max[a=#]>(d=Max[m=Tr/@(g=a[[#]]&/@Select[Subsets[Range[x=Length@#],2,x]&@#,FreeQ[Differences@#,1]&]&@a)]),Max@a,g[[#]]&@@@Position[m,d]]&

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Pyth, 19 bytes

esDm@LQdtf!q#1.+TyU

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esDm@LQdtf!q#1.+TyUQ Implicit: Q=eval(input())
 Trailing Q inferred
 UQ Generate range [0-len(Q))
 y Take the powerset of the above
 f Filter keep elements of the above, as T, using:
 .+T Take differences of consecutive elements of T
 q#1 Keep those differences equal to 1
 ! Logical NOT - empty lists evaluate to true, populated ones to false
 Result of the filter is those sets without consecutive numbers
 t Drop the first element (empty set)
 m Map the remaining sets, as d, using:
 L d For each element of d...
 @ Q ... get the element in Q with that index
 sD Order the sets by their sum
e Take the last element, implicit print

Gaia, 24 bytes

e:w;ċz⟨ọ1>¦ẏ⟩⁇‼⁇E‡ev2%Σ⌠

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Ugh, E‡ does some weird stuff...according to the documentation, it should do something like "given length i set of lists X and length j set of indices Y, return X[i][Y[j]]", but instead returns [X[i][Y[j]] X[i][Y[-j]] where negative indexing represents the complement, so we have to do ev2% to extract only the ones we want.

e				| eval as a list l
 :				| dup
 w				| wrap as a list
 ;				| push l again
 ċ				| push [1..len(l)]
 z				| push all subsets of [1..len(l)] -- index powerset.
 ⟨ ⟩⁇			| filter this for:
 ọ			| deltas
 1>¦			| are greater than 1
 ẏ			| all (all deltas greater than 1)
	 ‼⁇		| filter for non-empty lists
		 E‡		| table extract elements. Given l and index set i, this pushes
				| [l[i] l[setdiff(1..l,i)]] for some reason
		 ev2%		| get the l[i] only by unlisting, reversing, and taking every other element
		 Σ⌠	| Get the one with the maximum sum

J, 54 bytes

[:(>@:@/:+/&>)]<@#~"1[:(#~0=1 1+/@E."1])[:#:@.@i.2^#

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R, 108 107 bytes

function(v,M=-Inf)for(j in J<-seq(a=v))for(i in combn(J,j,,F))if(all(diff(i)>1)&sum(v[i])>sum(M))M=v[i]
M

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-1 thanks to @Giuseppe

Haskell, 300 168 bytes

import Data.List
h[]=1>2
h(x:y)=fst$foldl(a c->((fst a)&&(c-snd a>1),c))(1<2,x)y
z=snd.last.sortOn fst.map((,)=<<sum.snd).filter(h.fst).map unzip.subsequences.zip[0..]

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-132 bytes thanks to all the feedback from @nimi :)

Original

Ungolfed (original)

import Data.List
import Data.Function

f :: [Int] -> [(Int, Int)] -- attach indices for later use
f [] = []
f xs = zip xs [0..length xs]

g :: [[(Int, Int)]] -> [([Int], [Int])] -- rearrange into list of tuples
g [] = []
g (x:xs) = (map fst x, map snd x) : g xs

h :: [Int] -> Bool -- predicate that checks if the indices are at least 2 apart from each other
h [] = False
h (x:xs) = fst $ foldl (acc curr -> ((fst acc) && (curr - snd acc > 1), curr)) (True, x) xs
j :: [([Int], [Int])] -> [([Int], [Int])] -- remove sets that don't satisfy the condition
j xs = filter ((elements, indices) -> h indices) xs

k :: [([Int], [Int])] -> [(Int, ([Int], [Int]))] -- calculate some of elements
k xs = map ((elements, indices) -> (foldl1 (+) elements, (elements, indices))) xs

l :: [(Int, ([Int], [Int]))] -> ([Int], [Int]) -- grab max
l xs = snd $ last $ sortBy (compare `on` fst) xs

z -- put things together
```

Charcoal, 46 bytes

≔⟦υ⟧ηＦθ«≔υζ≔Ｅη⁺κ⟦ι⟧υ≔⁺ζηη»≔Φ⁺υηιη≔ＥηΣιζＩ§η⌕ζ⌈ζ

Try it online! Link is to verbose version of code. Explanation:

≔⟦υ⟧η

The variable u is predefined with an empty list. This is put in a list which is assigned to h. These variables act as accumulators. u contains the sublists that include the latest element of the input q while h contains the sublists that do not (and therefore are suitable for appending the next element of the input).

Ｆθ«

Loop over the elements of the input.

≔υζ

Save the list of sublists that contain the previous element.

≔Ｅη⁺κ⟦ι⟧υ

Take all of the sublists that do not contain the previous element, append the current element, and save the result as the list of sublists that contain the current element. (I don't use Push here as I need to clone the list.)

≔⁺ζηη»

Concatenate both previous sublists into the new list of sublists that do not contain the current element.

≔Φ⁺υηιη

Concatenate the sublists one last time and remove the original empty list (which Charcoal can't sum anyway).

≔ＥηΣιζ

Compute the sums of all of the sublists.

Ｉ§η⌕ζ⌈ζ

Find an index of the greatest sum and output the corresponding sublist.

Wolfram Language (Mathematica), 70 bytes

MaximalBy[Select[q=Rest@Subsets@#,Or@@(MatchQ[#~Riffle~_]/@q)&],Tr,1]&

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High-level search

 Select[q=Rest@Subsets@#, ] (*choose nonempty subsets of the input such that*)
 Or@@(MatchQ[ ]/@q)& (*there exists a subset of the input which matches*)
 #~Riffle~_ (*this list, with an item inserted between adjacent elements*)
MaximalBy[ ,Tr,1]& (*and return one with the greatest total*)

,1 is required so as not to inadvertently return invalid sets (otherwise, for example, an input of 1,1,1 would result in an output of 1,1,1,1,1,1)

Japt -h, 22 bytes

Êo à k_mÄ øZÃm!gU
fÊñx

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Japt -h, 21 bytes

Ever have one of those challenges where you completely forget how to golf?!

ð¤à fÊk_än ø1îmgUÃñx

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ð¤à fÊk_än ø1îmgUÃñx :Implicit input of array U
ð :Indices of elements that return true when
 ¤ : Converted to a base-2 string (to account for 0s)
 à :Combinations
 f :Filter by
 Ê : Length (to remove the empty combination)
 k_ :Remove elements that return true
 än : Deltas
 ø1 : Contains 1
 Ã :End remove
 ® :Map
 m : Map
 gU : Index into U
 Ã :End map
 ñ :Sort by
 x : Sum
 :Implicit output of last element

Python 2, 63 70 65 bytes

f=lambda a:a and max([a[:1],a[:1]+f(a[2:]),f(a[1:])],key=sum)or a

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5 bytes thx to ArBo