62nd Belarusian Mathematical Olympiad

a) It is impossible.

For $n=14$ we can divide the table into 49 of $2\times 2$ squares. If there are exactly 2 pawns in each of them, then the total number of the pawns in the table is equal to $2\cdot 49=98$. Since we have more than 98 pawns, we have $n>14$. On the other hand, if $n\ge 15$, then there are at least 105 pawns in the table (see solution of Problem A.8). It follows that 100 pawns cannot be arranged in the table.

b) It is possible.

The required arrangement of the 110 pawns for $n=15$ is given in the figure.