It is a well-known problem and many approaches are available on the web, look for Kadane's algorithm.

Below is a functional style implementation, see this link for the derivation

In[1]:= data = {1, 2, -4, 1, 3, -2, 3, -1};
        f[{y_, z_}, x_] := {Max[{0, x + y}], Max[Max[0, x + y], z]}
        Last[Fold[f, {0, 0}, data]]

Out[3]= 5

It is difficult to think of a solution that does not over-enumerate, except perhaps that no subsequence that begins with a negative number can be a candidate. So here is an exhaustive approach, which could be made much less understandable by more creative coding. ;-)

In[1]:= data = {1, 2, -4, 1, 3, -2, 3, -1};

In[2]:= (* start and end of all possible subsequences *)
indices = 
 Flatten[Table[{i, j}, {i, Length[data]}, {j, i, Length[data]}], 1]

Out[2]= {{1, 1}, {1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {1, 7}, {1, 
  8}, {2, 2}, {2, 3}, {2, 4}, {2, 5}, {2, 6}, {2, 7}, {2, 8}, {3, 
  3}, {3, 4}, {3, 5}, {3, 6}, {3, 7}, {3, 8}, {4, 4}, {4, 5}, {4, 
  6}, {4, 7}, {4, 8}, {5, 5}, {5, 6}, {5, 7}, {5, 8}, {6, 6}, {6, 
  7}, {6, 8}, {7, 7}, {7, 8}, {8, 8}}

In[3]:= subsequences = Take[data, #] & /@ indices

Out[3]= {{1}, {1, 2}, {1, 2, -4}, {1, 2, -4, 1}, {1, 2, -4, 1, 3}, {1,
   2, -4, 1, 3, -2}, {1, 2, -4, 1, 3, -2, 3}, {1, 2, -4, 1, 3, -2, 
  3, -1}, {2}, {2, -4}, {2, -4, 1}, {2, -4, 1, 3}, {2, -4, 1, 
  3, -2}, {2, -4, 1, 3, -2, 3}, {2, -4, 1, 3, -2, 3, -1}, {-4}, {-4, 
  1}, {-4, 1, 3}, {-4, 1, 3, -2}, {-4, 1, 3, -2, 3}, {-4, 1, 3, -2, 
  3, -1}, {1}, {1, 3}, {1, 3, -2}, {1, 3, -2, 3}, {1, 3, -2, 
  3, -1}, {3}, {3, -2}, {3, -2, 3}, {3, -2, 3, -1}, {-2}, {-2, 
  3}, {-2, 3, -1}, {3}, {3, -1}, {-1}}

In[4]:= sums = Total /@ subsequences

Out[4]= {1, 3, -1, 0, 3, 1, 4, 3, 2, -2, -1, 2, 0, 3, 2, -4, -3, 0, \
-2, 1, 0, 1, 4, 2, 5, 4, 3, 1, 4, 3, -2, 1, 0, 3, 2, -1}

In[5]:= (* large to small *)
orderingOfSums = Ordering[sums] // Reverse

Out[5]= {25, 29, 26, 23, 7, 34, 30, 27, 14, 8, 5, 2, 35, 24, 15, 12, \
9, 32, 28, 22, 20, 6, 1, 33, 21, 18, 13, 4, 36, 11, 3, 31, 19, 10, \
17, 16}

In[6]:= (* subsequence sum pairs *)
subsequencesAndSums = {subsequences[[orderingOfSums]], 
   sums[[orderingOfSums]]} // Transpose

Out[6]= {{{1, 3, -2, 3}, 5}, {{3, -2, 3}, 4}, {{1, 3, -2, 3, -1}, 
  4}, {{1, 3}, 4}, {{1, 2, -4, 1, 3, -2, 3}, 4}, {{3}, 
  3}, {{3, -2, 3, -1}, 3}, {{3}, 3}, {{2, -4, 1, 3, -2, 3}, 
  3}, {{1, 2, -4, 1, 3, -2, 3, -1}, 3}, {{1, 2, -4, 1, 3}, 
  3}, {{1, 2}, 3}, {{3, -1}, 2}, {{1, 3, -2}, 
  2}, {{2, -4, 1, 3, -2, 3, -1}, 2}, {{2, -4, 1, 3}, 2}, {{2}, 
  2}, {{-2, 3}, 1}, {{3, -2}, 1}, {{1}, 1}, {{-4, 1, 3, -2, 3}, 
  1}, {{1, 2, -4, 1, 3, -2}, 1}, {{1}, 1}, {{-2, 3, -1}, 
  0}, {{-4, 1, 3, -2, 3, -1}, 0}, {{-4, 1, 3}, 0}, {{2, -4, 1, 3, -2},
   0}, {{1, 2, -4, 1}, 
  0}, {{-1}, -1}, {{2, -4, 1}, -1}, {{1, 
   2, -4}, -1}, {{-2}, -2}, {{-4, 1, 3, -2}, -2}, {{2, -4}, -2}, {{-4,
    1}, -3}, {{-4}, -4}}

In[7]:= max = subsequencesAndSums[[1]]

Out[7]= {{1, 3, -2, 3}, 5}

Inspired by Gustavo's use of Partition (it works like a champ) ... here is the over-enumerated, but "functional paradigm", solution.

data={1,2,-4,1,3,-2,3,-1}; Max[Total /@ Flatten[Table[ Partition[data, i, j], {i, Length[data]}, {j, Length[data]}], 2]]

So, on this (puzzle) example, this solution (gets the right answer, 5) but it generates 121 subsequences in doing so. Using the procedural approach one need examine only a handful of subsequences though with lots of conditional statement executions. (Though I haven't, yet, formulated the worst-case for that approach.)

Still asking, is this - functional shortfall - intrinsic?

Thank you.

What are the rules for this problem? The way I read it, the answer is simply the sum of the positive entries (which would give the answer as 1+2+1+3+3). I am guessing that this is not the interpretation that you are using in getting the answer 1+3-2+3.