Divide the board into $4 \times 5$ boards.

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.