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.

1	8	2
6	9	5
3	7	4

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.