Clearly, a checkerboard pattern works. For the bound, split the board into $150^2$ $2 \times 2$ pieces. Since no three black cells form a corner, there is at most $2$ black squares per piece, since any three black tiles in a piece would create a corner. Thus, the bound is at most $2 \times 150^2$. This works for all $n$ for which $2 \mid n$