I use the user17410 equivalent formulation:
Input: $n$ vectors $X = \{ x_1, \dots, x_m \}$ over $\{0,1\}^n$, $n$ is part of the input
Question: Are there two different subsets $A,B \subseteq X$ such that
$$\sum_{x \in A} x = \sum_{x \in B} x$$
The hardness proof involve many intermediate reductions that follow the same "chain" used to prove the hardness of the standard EQUAL SUBSET SUM problem:
X3C $\leq$ SUBSET SUM $\leq$ PARTITION $\leq$ EVEN-ODD PARTITION $\leq$ EQUAL SUBSET SUM
(I'm still checking it so it may be wrong :)
STEP 1
The following problem (0-1 VECTOR SUBSET SUM) is NP-complete: given $X = \{ x_1, \dots, x_m \}$, $x_i$ vectors over $\{0,1\}^n$ and a target sum vector $t$, decide if there is $A \subseteq X$ such that
$$\sum_{x \in A} x = t$$
Proof: Direct reduction from EXACT COVER BY 3-SETS (X3C): given a set of $n$ elements $Y = \{y_1,...,y_n\}$ and a collection $C$ of $m$ three elements subsets $C = \{C_1,...,C_m\}$ we build the corresponding 0-1 VECTOR SUM instance setting $x_i[j] = 1$ if and only if element $j$ is included in $C_i$; $t = [1,1,...1]$.
STEP 2
Finding two equal sum subsets $A,B$ among $m$ 0-1 vectors over $\{0,1\}^n$, is equivalent to finding two equal sum subsets $A,B$ of vectors with element of bounded size $x_1 ... x_m$ where $max\{x_i\} = O((mn)^k)$ for fixed $k$.
For example the set of vectors:
x1 2 1 0 1
x2 1 2 3 1
Is equvalent to the 0-1 vectors:
x1 1 1 0 1 1 0 0 0 0
1 0 0 0 0 1 0 0 0
0 0 0 0 1 1 0 0 0
^ ^
+-- 0 elsewhere
x2 1 1 1 1 0 0 1 0 0
0 1 1 0 0 0 0 1 0
0 0 1 0 0 0 0 0 1
0 0 0 0 0 0 1 1 1
^ ^ ^
+-- 0 elsewhere
Informally the 0-1 vectors are grouped (if you select one vector of the x2 group and add
it to subset $A$, then you are forced to include in $A$ the other two and put the last in subset $B$) and the sums are done in unary (this is the reason why the corresponding non binary vectors must contain elements that are polynomially bounded with respect to $mn$).
So the following problem is NP-complete.
STEP 3
The following problem (0-1 VECTOR PARTITION) is NP-complete: given $B = \{ x_1, \dots, x_m \}$, $x_i$ vectors over $\{0,1\}^n$ decide if $X$ can be partitioned in two subsets $B_1, B_2$ such that
$$\sum_{x \in B_1} x = \sum_{x \in B_2} x$$
Proof: Reduction from 0-1 VECTOR SUM: given $X = \{ x_1, \dots, x_m \}$
and the target sum vector $t$; let $S = \sum x_i$, we add to $X$ the following vectors: $b' = -t + 2S$ and $b'' = t + S\;$: $B = X \cup \{b',b''\}$.
($\Rightarrow$) Suppose that there exists $A \subseteq X$ such that $\sum_{x \in A} x= t$;
we set $B_1 = A \cup \{b'\}$ and $B_2 =B \setminus B_1 = X \setminus \{A\} \cup \{b''\}$; we have
$$\sum_{x \in B_1} = b'+\sum_{x \in A} x = t - t + S = 2S$$
$$\sum_{x \in B_2} = b'' + \sum_{x \in X\setminus A} x = b'' + S - \sum_{x \in A} x=2S$$
($\Leftarrow$) Suppose that $B_1$ and $B_2$ have equal sum. $b', b''$ cannot both belong to the same set (otherwise their sum is $\geq 3S$ and cannot be "balanced" by the elements in the other set). Suppose that $b' = -t + 2S \in B_1$; we have:
$$-t +2S+ \sum_{x \in B_1 \setminus\{b'\}} x = t + S + \sum_{x \in B_2 \setminus\{b''\}} x$$
Hence we must have $\sum_{x \in B_1 \setminus\{b'\}} x = t$ and $B_1 \setminus\{b'\}$ is a valid solution for the 0-1 VECTOR SUM.
We only allow 0-1 vectors in the set $B$, so vectors $b', b''$ must be "represented
in unary" as shown in STEP 2.
STEP 3
The problem is still NP-complete if the vectors are numbered from $x_1,...,x_2n$ and
the two subsets $X_1,X_2$ must have equal size and we require that $X_1$ contains exactly one of $x_{2i-1},x_{2i}$ for $1 \leq i \leq n$ (so, by the equal size constraint, the other element of the pair must be included in $X_2$) (0-1 VECTOR EVEN-ODD PARTITION).
Proof:: The reduction is from 0-1 VECTOR PARTITION and is similar to the reduction from PARTITION to EVEN-ODD PARTITION. If $X = \{x_1,...,x_m\}$ are $m$ vectors over $\{0,1\}^n$ replace each vector with two vectors over $\{0,1\}^{2n+2m}$:
1 2 n
--------------------
x_i b_1 b_2 ... b_n
becomes:
1 2 ... 2i ... 2m
--------------------------
x'_2i-1 0 0 ... 1 ... 0 b_1 b_2 ... b_n 0 0 ... 0
x'_2i 0 0 ... 1 ... 0 0 0 ... 0 b_1 b_2 ... b_n
Due to the $2i$ element, the vectors $x'_{2i-1}$ and $x'_{2i}$ cannot be contained in the same subset; and a valid solution to the 0-1 VECTOR EVEN-ODD PARTITION correspond to a valid solution of the original 0-1 VECTOR PARTITION (just pick elements 2m+1..2m+n of each vector of the solution discarding vectors that contain all zeros in those positions).
STEP 4
0-1 VECTOR EQUAL SUBSET SUM (the problem in the question) is NP-complete: reduction from 0-1 VECTOR EVEN-ODD PARTITION similar to the reduction from EVEN-ODD PARTITION to EQUAL SUM SUBSET, as proved in Gerhard J. Woeginger, Zhongliang Yu, On the equal-subset-sum problem: given an ordered set $A = \{x_1,...,x_{2m}\}$ of $2m$ vectors over $\{0,1\}^n$, we build a set $Y$ of $3m$ vectors over $\{0,1\}^{2m+n}$.
For every vector $x_{2i-1}, 1 \leq i \leq m$ we build a vector $y_{2i-1}$ over $\{0,1\}^{2m+n}$ in this way:
1 2 ... i i+1 ... m m+1 m+2 ... m+i ... 2m 2m+1 ... 2m+n
------------------------------------------------------
0 0 ... 2 0 ... 0 0 0 1 0 x_{2i-1}
For every vector $x_{2i}, 1 \leq i \leq m-1$ we build a vector $y_{2i}$ over $\{0,1\}^{2m+n}$ in this way:
1 2 ... i i+1 ... m m+1 m+2 ... m+i ... 2m 2m+1 ... 2m+n
------------------------------------------------------
0 0 ... 0 2 ... 0 0 0 1 0 x_{2i}
We map element $x_{2m}$ to
1 2 ... ... m m+1 m+2 ... . 2m 2m+1 ... 2m+n
------------------------------------------------------
2 0 ... ... 0 0 0 1 x_{2m}
Finally we add $m$ dummy elements:
1 2 ... ... m m+1 m+2 ... ... 2m 2m+1 ... 2m+n
------------------------------------------------------
4 0 ... ... 0 0 0 0 0 ... 0
0 4 ... ... 0 0 0 0 0 ... 0
...
0 0 ... ... 4 0 0 0 0 ... 0
Note again that vectors containing values $> 1$ can be represented in "unary" using a group of 0-1 vectors like showed in STEP 2.
$Y$ has two disjoint $Y_1,Y_2$ subsets having equal sum if and only if $X$ has an even-odd partition.
