I'm not sure if 4 is possible, but I thought that I'd try it the math way before I tried to write code to brute force the solution.

First, I want to see if 64 square coverage is possible with 4 queens. First, if you place one queen on a center square, it covers:

1 (it's own square, based on original problem)

+ 7 (row)

+ 7 (column)

+ 7 (diagonal 1)

+ 6 (diagonal 2)

= 28 squares.

Now, if you could keep placing Queens on the board, each covering 28 squares, then you could cover 103 points with 4 queens, but this is not the case.

First, we have to look at the Queen as a rook. If you place two rooks on two points on the board, and made sure they were neither in the same row or the same column, together they would double cover one point. Thus, the effectiveness of each subsequent rook is diminished by 2*m, where m = the number of rooks previously placed on the board.

Next, due to the nature of the Queen as a bishop, we have to look at centralization. The above sum is the result of the Queen being centralized, meaning, in the center 4 squares. The next Queen that is placed CANNOT be in the center four squares, because of what I would like to call 'heavy collision', where:

collision (c) = the number of squares double protected by any set of two queens on the board.

Now, collision, for two opposite colored Queens near the center of the board (one at e4, one at c5, if you will), is 10! This is the problem that will kill us in the end. The Second placed Queen will reduce in effectiveness from not only the fact it is in the second row from the center, where it's effectiveness is reduced, but also collision.

On that previous statement, the effectiveness of a Bishop as it moves away from the center of the board, is diminished by a count of 2. Thus, the function b(x) where x = the distance from the center for x >= 0 , x <= 3 ranges from 13 to seven. We also have to add that the more centralized bishops of opposite color are in relation to a rook the more collision occurs, in the order of 6 at a maximum (see the collisions of the previous example) and likely 2 at a minimum. Let us just use these values and put in the possibilities from Q1 to Q4:

Q1 = 28

Q2 = 16

Q3 = 16 - 4 - 2 = 10 (I'm assuming collision reduction

Q4 = 16 - 4 - 2 - 2 = 8 (Assuming Collision reduction)

28 + 16 + 10 + 8 = 62.

I would like to see if even this number is possible. I've gotten 59 square coverage with 4 Queens (e4, c5, f6, d2) and a fifth Queen would finish the puzzle (b3).

While my math is SHAKY and heavily generalized, it is good enough, I think, to prove it is impossible. Even as the collision with the central Queens diminishes, other things subtract enough from the placed Queens to make it impossible.