, 2015

We assume each nonblack square is white. Repeat the $3 \times 3$ block with black squares down the main diagonal $9$ times. This is a possible construction with $3 \times 9 = 27$ black squares. We shall prove that this is the only answer.

Suppose there are $26$ black squares. We name the bottom left square as $(1, 1)$. Remove the squares in the block $\{(1, 1), (1, 2), (1, 3)\}$. The remaining squares can be partitioned into $3 \times 2$ or $2 \times 3$ blocks which contain $26$ black squares. Thus the removed squares are white. Repeat the argument with the blocks $\{(3, 1), (3, 2), (3, 3)\}$, $\{(1, 1), (2, 1), (3, 1)\}$, $\{(1, 3), (2, 3), (3, 3)\}$, we conclude that all these squares are white. Thus we have a $3 \times 3$ block with at most $1$ black square (see figure below). This is a contradiction.

Suppose there are $28$ black squares. Using the same argument as above, we conclude that there are $2$ black squares in each of the blocks $\{(1, 1), (1, 2), (1, 3)\}$, $\{(3, 1), (3, 2), (3, 3)\}$, $\{(1, 1), (2, 1), (3, 1)\}$ and $\{(1, 3), (2, 3), (3, 3)\}$. Thus all the squares in the three blocks $\{(2, 1), (2, 2), (2, 3)\}$, $\{(1, 2), (2, 2), (3, 2)\}$, $\{(1, 4), (2, 4), (3, 4)\}$ are white as shown in the following diagram, again leading to a contradiction.