Solution:
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 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. This is indeed attainable as in the diagram below.