Second Round of the 73rd Czech and Slovak Mathematical Olympiad (January 16th, ), 2024

a) Yes, the following table gives us the desired quadruple of sums.

b) No, it is impossible to get this quadruple. Denote the sum of the four numbers as $S$, we shall proceed by proving that $S \le 98$ and then describing all the cases where we get an equality.

Note that in $S$, the central number contributes four times, the four corner tiles contribute once and the remaining four tiles contribute twice, hence we get:
$$S \le 4 \cdot 9 + 2 \cdot (8 + 7 + 6 + 5) + (4 + 3 + 2 + 1) = 98.$$
Moreover, it is clear that the upper bound is attained if and only if the number $9$ is in the centre and numbers $1$, $2$, $3$, $4$ are in the corners in some order.

Suppose that we were able to obtain the quadruple $20$, $23$, $26$, $29$, since the sum of the four numbers is $98$, the placement of the numbers in the big table would have to follow the rules explained above. But this means that the smallest possible sum of a $2 \times 2$ square is $1 + 5 + 6 + 9 = 21$, a contradiction, hence we can't obtain this quadruple.