Construction of binary puzzle
To meet condition 1, $n$ must be even, so let's assume that it is. Then we can automatically achieve both conditions 1 and 2 by making an $n/2\times n/2$ matrix whose entries are $2\times 2$ submatrices in one of the two patterns
$$ \left( \begin{array}{cc} 0 &amp; 1 \\ 1 &amp; 0 \end{array} \right),\quad \left( \begin{array}{cc} 1 &amp; 0 \\ 0 &amp; 1 \end{array} \right), $$
as these prevent any three consecutive digits (in either direction) from being equal.

Now all you need to do is to choose which of these two $2\times 2$ matrices to use in which position of the $n/2\times n/2$ matrix. You must now avoid both equal rows and columns, and complementary rows and columns, but that's still very easy, for instance by putting the 1-0-0-1 matrices down the main diagonal and the 0-1-1-0 matrices everywhere else.