Whichet

Look at the following image, where the red are the non-empty squares. Non-empty squares

• Because the red squares come in pairs, each flea in a red square can jump to the other red square (and not to an empty square!).
• Notice that each empty square is adjacent to at least one red square, so each flea in an empty square can jump to a non-empty red square.
• I believe the solution (coverage) is minimal because each non-empty square has exactly and only 1 red neighbor.
• There are 20 red squares, so $\boxed{44}$ squares after the jump.

Jumps

Main idea : Find a lower bound on the number of squares occupied and realize it by a model.

Complete proof :

First, color the board in a chessboard pattern. Observe that fleas on the black squares will end up on white squares and vice versa, and they don't disturb each other, so our task is simplified to observe only the fleas on the black squares that moves to white squares and multiply the result by two.

Next, observe the following image . Light blue squares are black ones. Note that no white square is adjacent to more than two white circles, so no three fleas from white-circled squares can gather in a single white square. There are 20 white circles, so we need at least 10 occupied white cells. This is also realized by the black circles; note that every light blue square is adjacent to at least one black circle, so they can move there. This frees up $32 - 10 = 22$ white squares.

By symmetry, we can also free up $22$ black squares, so the maximum number of squares freed is $22+22 = \boxed{44}$ .

Motivation : To be honest, I'm not sure how I came up with that. I quickly obtained the model with $10$ filled squares easily by doing some greedy collections, but it's surprisingly hard to prove that. I tried to put the white circles so no white square is adjacent to more than one white circle, but I kept only getting $9$ on the board. So I somehow got the idea to use two white circles per white square, and for some reason got the above quickly. I'm still figuring out whether the maximum number of white circles such that no white square is adjacent to more than one white circle is indeed $9$ or not; if it is, then this problem is one of the most devious problems I've ever seen. If it isn't, then it's still one of the most devious since then the configuration is hard to obtain.

Generalization : Making it to any board size of even length seems not terribly fun; that configuration is easily extensible. More interesting is of odd length, as the symmetry of white and black squares is broken. I believe this deserves its own discussion...

Since each flea starting on a white square jumps to a black square and vice versa, we can consider the fleas starting on black separately to the fleas starting on white. In fact, if we find an optimal solution for the fleas starting on black, then by symmetry that will give us the optimal solution for the fleas starting on white.

I'm going to use chess notation here for simplicity. To maximise the number of empty white squares after the whip, it is equivalent to minimizing the number of kings we can place on white squares such that every black square is attacked (i.e. is adjacent to a king). The squares with kings correspond to the squares that fleas jump to, and any flea is adjacent to a king because we stipulated that.

Here is a solution with $10$ kings: the kings are on d1, a2, g2, d3, b5, f5, h5, d7, a8, g8.

Now we will show that $9$ kings is not possible, by showing that in the lower-left half of the board, there must be $4$ kings which only cover at most $13$ squares between them. The same argument will also show that the upper-right half of the board must have the same; so the maximum coverage of $9$ kings will be $13 + 13 + 4 = 30$ , but there are $32$ black squares.

Consider the square $a1$ . To attack it, there must be a king on $a2$ or $b1$ . By symmetry we can assume this king is on $a2$ without loss of generality.

Now consider $c1$ . To attack this square there must be a king on $c2$ or $d1$ . (Having a king on $b1$ is sub-optimal, this will become clear shortly if it isn't already).

In either case, only three new squares are attacked, so we have $2$ kings placed with only $6$ squares attacked.

If the second king is on $c2$ , then consider the square $e1$ . Regardless of which square a king is placed on to attack $e1$ , it can only attack $3$ new squares. So we would have $3$ kings placed that only cover $9$ squares, therefore (choosing any other king in this half) there are $4$ kings that only cover $13$ squares.

If, instead, the second king is on $d1$ , consider the square $b4$ . In order to place a third king that attacks at least $3$ new squares including this one, that king must be on $b5$ or $c4$ . In both cases, the third king attacks $4$ new squares.

If the third king is on $b5$ then consider $c3$ . Regardless of where a king is placed adjacent to $c3$ , at most one new square is attacked besides $c3$ , so our first four kings attack at most $12$ squares.

If the third king is on $c4$ then consider $a5$ . Regardless of where a king is placed adjacent to $a5$ , at most $3$ new squares are attacked including $a5$ , so our first four kings attack at most $13$ squares.

We have exhaustively traversed a tree of possibilities now, and proven that $9$ kings can only cover $30$ squares.

Therefore $10$ kings must be on white squares. Going back to the fleas, this means there can be at most $32 - 10 = 22$ empty black squares after the whip. By symmetry there can be at most $22$ empty white squares, for a total of $22 + 22 = \boxed{44}$ .