(test final). The minimum number of free cells is 99.

In this paragraph, we will see how to leave at most 99 free cells. Among the South-East diagonals, 99 have an odd number of cells. On each such directed diagonal, we group the flies into pairs of adjacent flies. At most one fly remains alone (in the case of an odd length). The adjacent flies can swap places. We let the "single" flies go wherever they want. The only flies that can leave their place vacant are the "single" flies, which number 99 (one per diagonal of odd length). We therefore know that the sought number is at most 99.

There cannot be fewer than 99 free cells. Indeed, number the columns from 1 to 99 and imagine that the even-numbered columns are blue and the others are red. We say that a fly is of a certain color if it is initially on a column of that color. Since a move sends each fly to a cell of a different color, the only flies that can try to cover the $50 \times 99$ red cells are the $49 \times 99$ blue flies. There will therefore be $99 \times (50-49) = 99$ free red cells.
. We have a checkerboard with a side length of 99. There is a fly on each square. In one step, each fly must move exactly one square diagonally (several flies can then end up on the same square). After one step, what is the minimum number of empty squares?