**TEMPORARILY WITHDRAWN.** Oleg567 has correctly noted that at least one of these is in error.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. ---------- 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.