Problem 1871

Solution:

The maximum and the minimum are, respectively, $30$ and $19$ and are realized, for example, by the following tables

Indeed, in order for the sum of the $9$ digits in a square $3 \times 3$ to be divisible by $4$, there are only two possibilities: to have $6$ times the digit $1$ and $3$ times the digit $2$ (sum $12$), or to have $2$ times the digit $1$ and $7$ times the digit $2$ (sum $16$).

It is therefore evident that in the $4 \times 4$ table there are at least $2$ digits $1$. If we arrange them, in whatever way, inside the central $2 \times 2$ square of the table, that is, so as to be contained in all $4$ squares $3 \times 3$, no further digits $1$ are necessary. The maximum sum is therefore realized with $2$ digits $1$ and $14$ digits $2$, for a total of $30$.

Similarly, in the $4 \times 4$ table there are at least $3$ digits $2$, and no further ones are needed if these are arranged, in whatever way, inside the central $2 \times 2$ square of the table. The minimum sum is therefore realized with $3$ digits $2$ and $13$ digits $1$, for a total of $19$.