Solution:
The largest achievable score is 17, as shown by the example pictured. It remains to show that the maximum score of 18 cannot be achieved. To this end, we assume the existence of a placement achieving 18 points and first consider the rows and columns. Neither in a row nor in a column can there then lie an odd number of pieces, so that, because of the sum 10, only (up to permutations) the possibilities (4;2;2;2) or (4;4;2;0) can occur. However, if there is a row (column) with 0 pieces, then all columns (rows) can contain at most 2 pieces because of the required evenness, so that the sum 10 is not reached. Therefore there is exactly one row and one column with 4 pieces. The intersection point of these

17 points

Case a)

Case b)

Case c)
two lines can now lie a) in a corner, b) on the edge, but not in a corner, c) in the interior of the square. Because of symmetry, it suffices to consider one example each.
Case a): Here the five NW-SE lines each contain, at this point, only one piece. But since only three more pieces are to be placed, the score of 18 can no longer be achieved.
Case b): Here four of the NW-SE lines each contain, at this point, only one piece. But since only three more pieces are to be placed, the score of 18 can no longer be achieved.
Case c): The two NW-SE lines with exactly 2 squares each contain, at this point, only one piece. If a second piece is placed on them, then the upper SW-NE line with 3 pieces scores no point. Here too the score of 18 is not achieved.