Estonian Mathematical Olympiad

There are $8$ possible sums: $$\begin{array}{ccc}1+5+9,& 2+6+7,& 3+4+8,\\ 1+6+8,& 2+4+9,& 3+5+7,\\ 2+5+8,& 4+5+6.& \end{array}$$ Note that no two of the three sums in the first row have any common addends, and the same applies to the three sums in the second row. Moreover, each sum in the first two rows has one common addend with every sum outside its own row, including the sums in the third row. Thus, Pets can choose one of the first two rows (one from which Ats did not choose a sum in the first move) and write any sum from it on his turn. This guarantees a common addend with the sum that Ats has written on his turn. Since Ats cannot write a sum from the same row from which Pets has chosen his sum on his second move, Pets can choose any unused sum from that row on his second move, because it has a common addend with the sum written by Ats and no common addends with the sum last written by Pets. Analogously, Pets can make his third move if Ats has not already lost by then.

When Pets has made his third move, there are only two sums left unused. This means that if Ats were to make another move, one of the sums in the third row should have been written on the board either on this move or on an earlier move. According to Pets' strategy, Ats had to do this. However, the sums in the third row have one common addend with every other sum, so according to the rules they cannot have either the second previous sum or the sum after the next sum. This contradicts the assumption that the game can continue after Pets' third move. So in fact, Ats has no more suitable sums left after Pets' third move and has lost.