XXVII Olimpiada Matemática Rioplatense

For $i=1,\dots ,4$, let $a_i$ be the maximum number in column $i$, and let $b_1,b_2,\dots ,b_{12}$ be the remaining 12 numbers written on the board (different from $a_1,a_2,a_3,a_4$). Then, for every $i$, $a_i$ is the sum of the other three numbers in column $i$; therefore, $$a_1+a_2+a_3+a_4=b_1+b_2+\cdots +b_{12}.(2)$$ Since $b_1,b_2,\dots ,b_{12}$ are different positive integers, we have that $$b_1+b_2+\cdots +b_{12}\ge 1+2+\cdots +12=78.(3)$$ On the other hand, since $a_1,a_2,a_3,a_4$ are also different positive integers, and $M$ is the largest number on the board, then $$a_1+a_2+a_3+a_4\le M+(M-1)+(M-2)+(M-3)\le 4M-6.(4)$$ From (2), (3) and (4), it follows that $$4M-6\ge a_1+a_2+a_3+a_4=b_1+b_2+\cdots +b_{12}\ge 78,$$ which implies that $M\ge 21$. The following is an example with $M=21$:

1	8	12	21
7	9	20	4
10	19	3	6
18	2	5	11