It's enough to choose the following numbers: $-2$, $-1$, $0$, $1$, $2$. Then we can write down the sets of integers, but we can also apply the following reasoning: for any two numbers, say, $a$, $b$, selected by Petrik, from one side exists pair of numbers $(-a, -b)$, whose sum is the opposite to the initial, and, from another side, there exists a triple of numbers except $a$, $b$, and their sum also is $(-a - b)$, as the sum of all five numbers is zero. So we get a correspondence between the numbers from the left and from the right parts of the board.