Subset Sum Variant still NP hard?
Note that $T=S$ and $T=\emptyset$ are trivial solutions.

With the constraints $T \neq \emptyset$ and $S\setminus T \neq \emptyset$ the problem seems (unless the proof below is wrong :-) NP-complete. A possible reduction is from PARTITION: given a set $A$ of $n$ positive integers $A=a_1,..a_n$, can $A$ be partitioned in two sets $A = A_1 \cup A_2$ such that $\sum_{a_i\in A_1} a_i = \sum_{a_j \in A_2} a_j$?

Let $M = \sum_{a_i \in A}a_i$; without loss of generality we can assume that $M$ is even.

You can pick $S = A \cup \{a_{n+1}, a_{n+2}\}$ with $a_{n+1}= a_{n+2}=-M/2$; so we have $\sum_{s \in S}s = 0$ (right hand side of the equation of your problem). 

The original PARTITION problem has a solution if and only if your problem has a solution.

$(\Leftarrow)$ Suppose that there exists $T$ such that $|\sum_{t\in T}t-\sum_{s\in S\setminus T}s|\le\sum_{s\in S}s$ = 0. So the absolute value of the difference of the sums is zero: $|\sum_{t\in T}t-\sum_{s\in S\setminus T}s| = 0$  $T$ cannot contain both $a_{n+1}$ and $a_{n+2}$, otherwise  $|\sum_{t\in T}t-\sum_{s\in S\setminus T}s| &gt; 0$ because at least one element $a_i&gt;0$ must be contained in $S \setminus T$ and $|-M/2 -M/2 - a_i| &gt; M$. For the same reason $S \setminus T$ cannot contain both $a_{n+1}$ and $a_{n+2}$. So $a_{n+1} \in T$ and $a_{n+2} \in S \setminus T$, and their contribution to the absolute value of the difference of the sums is $0$. So the remaining elements must be partitioned in such a way that the sum of the elements in $T$ equals the sum of elements in $S \setminus T$ in order to have an absolute value of the difference of the sums equal to $0$.

$(\Rightarrow)$ If the original partition problem has a solution, then it is easy to see that if $T = A_1 \cup a_{n+1}$ we get an absolute value of the difference of the sums equal to $0$.