I get $18$ sum-free binary vectors recently.
(UPDATE: It is not top-result now. Brendan McKay obtained $\large\bf 19$ sum-free binary vectors. See his answer, update Feb 24.)
A few examples of $18$ sum-free binary vectors:
$\qquad(0,0,0,0,1,1,0,0,1,1)$,
$\qquad(0,0,0,1,1,0,1,0,0,1)$,
$\qquad(0,0,1,1,0,1,0,0,1,1)$,
$\qquad(0,0,1,1,1,1,0,1,1,0)$,
$\qquad(0,1,0,0,1,1,1,0,1,0)$,
$\qquad(0,1,0,1,0,0,0,1,0,0)$,
$\qquad(0,1,0,1,0,0,1,1,1,0)$,
$\qquad(0,1,0,1,1,0,0,0,0,1)$,
$\qquad(0,1,1,0,1,0,0,1,0,1)$,
$\qquad(0,1,1,1,0,1,1,1,0,1)$,
$\qquad(1,0,0,0,1,0,1,1,0,1)$,
$\qquad(1,0,0,0,1,0,1,1,1,1)$,
$\qquad(1,0,0,1,0,0,0,1,0,1)$,
$\qquad(1,0,1,0,0,1,0,1,0,1)$,
$\qquad(1,1,0,0,0,1,1,0,0,1)$,
$\qquad(1,1,0,0,1,0,0,0,1,1)$,
$\qquad(1,1,0,1,1,1,0,0,0,0)$,
$\qquad(1,1,1,0,1,0,1,0,0,0)$;
$\qquad(1,0,0,0,1,1,1,1,0,1)$,
$\qquad(0,1,0,1,0,0,1,0,1,0)$,
$\qquad(0,1,1,0,1,0,0,1,0,1)$,
$\qquad(1,1,0,0,1,0,0,0,1,1)$,
$\qquad(1,1,0,1,1,0,1,1,1,0)$,
$\qquad(0,1,0,1,0,0,0,1,0,0)$,
$\qquad(0,0,1,1,0,1,0,0,1,1)$,
$\qquad(1,1,0,0,0,1,1,0,0,1)$,
$\qquad(0,1,0,0,1,1,1,0,1,0)$,
$\qquad(0,1,0,1,0,0,1,1,1,0)$,
$\qquad(1,0,1,0,0,1,0,1,0,1)$,
$\qquad(1,1,1,0,0,0,1,0,0,0)$,
$\qquad(0,0,0,0,1,1,0,0,1,1)$,
$\qquad(1,1,1,0,1,0,1,0,0,0)$,
$\qquad(1,0,0,1,0,0,0,1,0,1)$,
$\qquad(1,1,0,1,1,1,0,0,0,0)$,
$\qquad(1,0,1,0,1,1,0,1,0,0)$,
$\qquad(0,0,0,1,1,0,1,0,0,1)$.
(Here is discussion of sum testing - as wide comment for Brendan McKay)
Testing of all sums is slow, when you'll generate all sums, and compare each-other. If there are $n$ vectors, then there are $N=2^n$ sums. Naive comparison will take $O(N^2)=O(2^{2n})$ of time.
Good way to construct all sums:
shown on an example of $5$ vectors $a,b,c,d,e$.
There are $2^5=32$ sums.
1-st step: Constructing of sums:
- $s[0]=0$;
- $s[1]=a$;
- $s[2]=b+s[0]$, $s[3]=b+s[1]$;
- $s[4]=c+s[0]$, $s[5]=c+s[1]$, $s[6]=c+s[2]$, $s[7]=c+s[3]$;
- $s[8+i]=d+s[i]$, where $i=0,1,...,7$;
- $s[16+i]=e+s[i]$, where $i=0,1,...,15$.
Time capacity is $O(n\times N) = O(n\times 2^n)$.
2-nd step: Sorting of sums:
Fast sorting methods are quick-sort, heap-sort etc...
Time capacity is $O(N \log N) = O(2^n \times n)$.
3-nd step: Comparison of neighboring sums:
for each $i = 1,..,N-1$ just compare $s[i-1]$ and $s[i]$.
Time capacity is $O(N) = O(2^n)$.
So, total time capacity is $O(n\times N) = O(n\times 2^n)$. Much faster than $O(N^2)$.