We are asked to find all natural numbers $n$ for which we can paint the cells of a $2n \times 2n$ chessboard black and white such that:
- each pair of cells symmetric about the center of the board consists of two different colored cells;
- each row/column has exactly $n$ black and $n$ white cells.
Claim: the answer is all even $n$, because as I have observed
- for even $n$ we can paint the upper half of the board in a standard way and then reflect it about the center of the board and that (hopefully) will get us the right coloring. I need help proving the construction works for even $n$.
- for odd $n$ this construction fails, however it gives us a hint, because if we say we have $k_1$ black cells in the upper half of the first column and $k_{2n}$ black cells in the upper half of the last column then by symmetry $k_1 + (n - k_{2n}) = n$, so $k_1 = k_{2n}$, but this is impossible as by symmetry $k_1 = n - k_{2n}$. Is it sufficient?
To prove it is impossible for odd $n$, let $q_1,q_2,q_3$ and $q_4$ be the number of black squares in each quadrant of the board, with the indices chosen so $q_{i+1}$ and $q_i$ refer to adjacent quadrants, for each $i\in \{1,2,3\}$. The conditions on row sums and column sums imply $$ q_1+q_2=q_2+q_3=q_3+q_4=q_4+q_1=n^2 $$ Furthermore, the symmetry condition implies $$ q_1+q_3=q_2+q_4=n^2 $$ These equations combine to prove that each $q_i=n^2/2$, which is a contradiction.