In chess, a normal knight goes two steps forward and one step to the side, in some orientation. Thanic thought that he should spice the game up a bit, so he introduced a new kind of piece called a warrior. A warrior can either go three steps forward and one step to the side, or two steps forward and two steps to the side in some orientation. In a $2020 \times 2020$ chessboard, prove that the maximum number of warriors so that none of them attack each other is less than or equal to $\frac{2}{5}$ of the number of cells.
Source: BdMO 2019 National Secondary P10
We divide the board into $4 \times 5$ boards and prove that each of them has at most $\frac{2}{5}\cdot 20 = 8$ warriors. From that, it will follow that the whole board has at most two fifth warriors.
In the figure, in the cells of the same color, there can be at most $1$ warrior. So, there can be at most $8$ warriors in a $4 \times 5$ board.
Similarly, it is true for $2020 \times 2020$ because the $2020 \times 2020$ board can be divided into $4 \times 5$ boards.
Divide the board into $4 \times 5$ boards.
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