**NEW VERSION** 
The earliest version of this answer was incorrect, as noted by Rob Pratt and Oleg567. My new code was apparently correct but there was an embarrassing bug in old code used with it. 
Keep fingers crossed...

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Call two solutions equivalent if one is obtained from the other by permuting columns.  I made some crude code, not well checked, and found these optimal solutions for vectors of length $n$.

$n=2$: best is 2, 2 nonequivalent solutions, example

    10
    11

$n=3$: best is 4, with 2 nonequivalent solutions, example

    110
    101
    011
    111

$n=4$: best is 5, with 48 nonequivalent solutions, example

    1100
    1011
    0111
    1110
    1111

$n=5$: best is 7, with 877 nonequivalent solutions, example

    11100
    11010
    11001
    10111
    01111
    11110
    11111

$n=6$: best is 9, with 114227 nonequivalent solutions, example

    111000
    100111
    010111
    001111
    110110
    111100
    111011
    111110
    111101

$n=7$: best is 12, with 118485 nonequivalent solutions, example

    1000000
    0100000
    0010000
    1001000
    1101100
    1001011
    0011110
    1110010
    0111010
    0111001
    0100111
    1010101

The method I'm using might be able to do $n=8$, but $n=10$ is impossible. If someone could verify the above are solutions, that would improve confidence in the results.

The slowest part by far is testing for equal subset sums.  I run through all the subsets using a gray code, with only a few machine instructions for each. But what it really needs is some way to test the condition without enumerating subsets.  Is there one?

It took about 45 seconds to do $n=6$ and 220 hours to do $n=7$.  All the above solutions can be found on my [combinatorial data page][1].

**ADDED** Feb 24: For $n=8,9,10$ I have not proved what the largest size is.  It would be plausible to prove it for $n=8$ but I don't know how to do larger sizes.

For $n=8$, I have more than 4 million sets of size 14 and there are more.

For $n=9$, I have 160445 sets of size 16 but there are more.

For $n=10$, I have 2999 sets of size 19 and will find more.  Here is one:

    0001111011
    1010111100
    1011010001
    0111010100
    1110110111
    1101000101
    1011100010
    0100110011
    0110010001
    0101101100
    1101101101
    0110100011
    0001010110
    0000110110
    1100011010
    1000101001
    0010001111
    1000010111
    0110001000

  [1]: http://cs.anu.edu.au/~bdm/data/dissociated.html