Problem 1131

Let $n$ be an odd number and $a_1 < a_2 < \dots < a_n$ 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. ■