**TEMPORARILY WITHDRAWN.**  
Oleg567 has correctly noted that at least one of these is in error.  I will track down the program bug and do the computations again.  Meanwhile, don't take these "solutions" seriously.

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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
    01

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

    110
    101
    011
    111

$n=4$: best is 6, 5 nonequivalent solutions, example

    1100
    1010
    1001
    0110
    1110
    1101

$n=5$: best is 8, 265 nonequivalent solutions, example

    11000
    10110
    10101
    01011
    01110
    11100
    11101
    01111

$n=6$: best is 10, 80066 nonequivalent solutions, example

    100000
    010000
    111000
    100110
    001110
    010111
    101101
    101011
    101111
    110111

$n=7$: there are examples of size 13, not proved maximal yet, example

    1000000
    0100000
    0010000
    1101100
    1100011
    1011010
    0001111
    1010101
    0110110
    1110100
    0111001
    1111011
    1101111

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.