, 2015

If the chessboard is colored black and white as usual, then any diagonal is a solid color. So we may consider bishops on black and white squares separately.

In one direction, the lengths of the black diagonals are $2, 4, 6, 8, 6, 4$, and $2$. Each of these can have at most three bishops, except the first and last diagonals which can have at most two, giving a total of at most $2 + 3 + 3 + 3 + 3 + 3 + 2 = 19$ bishops on black squares. Likewise there can be at most $19$ bishops on white squares for a total of at most $38$ bishops.

Conversely, if we place $38$ bishops on the four boundaries of the table and on the second and seventh rows except the second and seventh square of the second row, as shown in the picture, one can check that this arrangement satisfies the condition of the problem.