Iranian Mathematical Olympiad

Let $n$ be an odd number and $a_1<a_2<\cdots <a_n$ be the given real numbers. Put $a_i-a_j$ (where $i>j$) into the first set if $i$ and $j$ have the same parity and put them into the second set if they have different parity. We claim that

the coefficient of $a_i$'s in both sets are the same. Suppose $i$ is even (the second case is similar). The first set contains the elements $$a_{n-1}-a_i,a_{n-3}-a_i,\dots ,a_{i+2}-a_i,a_i-a_{i-2},a_i-a_{i-4},\dots ,a_i-a_2,$$ that have $a_i$ appeared in them. Thus the coefficient of $a_i$ in the first set is $$\frac{i-2}{2}-\frac{n-i-1}{2}=\frac{2i-n-1}{2}$$ Similarly, the second set contains the elements $$a_n-a_i,a_{n-2}-a_i,\dots ,a_{i+1}-a_i,a_i-a_{i-1},a_i-a_{i-3},\dots ,a_i-a_1$$ that also have $a_i$ appeared in them. So the coefficient of $a_i$ in the second set is $$\frac{i}{2}-\frac{n-i+1}{2}=\frac{2i-n-1}{2},$$ Hence $a_i$'s coefficient in all member's sum of both sets are same and we are done. ■