Problem 1679

By Taylor's theorem, we have
$$\sqrt{1-t} = 1 - \sum_{k=1}^{\infty} \frac{(2k)!}{4^k (k!)^2 (2k-1)} t^k$$
for any $-1 < t < 1$. Since $\sum_{i=1}^n x_i = 0$, we have
$$\sum_{i=1}^{n} \sum_{j=1}^{n} x_i x_j \sqrt{1 - x_i^2 x_j^2} = \left( \sum_{i=1}^{n} x_i \right)^2 - \sum_{k=1}^{\infty} \frac{(2k)!}{4^k (k!)^2 (2k-1)} \left( \sum_{i=1}^{n} x_i^{2k+1} \right)^2 \le 0.$$
Equality holds if and only if $\sum_{i=1}^n x_i^{2k+1} = 0$ for all $k \ge 1$. This means that the list $x_1, x_2, \dots, x_n$ consists of zeros and opposite numbers. Indeed, equality holds for such numbers. Now suppose $\sum_{i=1}^n x_i^{2k+1} = 0$ for all $k \ge 1$. Removing the zeros, and negating the negative numbers, we get positive numbers $a_1, a_2, \dots, a_m > 0$ and $b_1, b_2, \dots, b_l > 0$ such that $\sum_i a_i^{2k+1} = \sum_j b_j^{2k+1}$ for all $k \ge 1$. It is easy to see that this implies $\max\{a_i\} = \max\{b_j\}$ considering a large enough $k$. By induction, the list consists of zeros and opposite numbers.