Solution:
Let the given numbers (in an arbitrary order) be $b_{1}, b_{2}, \ldots, b_{n}$. For every possible permutation $\pi$ of $\{1,2, \ldots, n\}$, consider the sum
$$b_{\pi(1)} b_{\pi(2)}+b_{\pi(2)} b_{\pi(3)}+\cdots+b_{\pi(n-1)} b_{\pi(n)}+b_{\pi(n)} b_{\pi(1)} .$$
We wish to show that some such sum is negative, so assume otherwise. For every two distinct elements $i, j \in\{1, \ldots, n\}$, the term $b_{i} b_{j}$ appears $N$ times among these sums, where $N$ does not depend on $i, j$, by symmetry. (In fact, one can show that $N=n(n-2)$ !.) Each such sum is assumed to be nonnegative; adding these inequalities for all permutations $\pi$, and dividing by $N$, we have
$$\sum_{i \neq j} b_{i} b_{j} \geq 0$$
However, we also know that $\sum_{i} b_{i}^{2}>0$ (strictly, since not all $b_{i}$ are zero). Thus
$$\left(b_{1}+\cdots+b_{n}\right)^{2}=\sum_{i} b_{i}^{2}+\sum_{i \neq j} b_{i} b_{j}>0$$
But since $b_{1}+\cdots+b_{n}=0$, we have a contradiction. So our assumption was false, and the needed negative sum does exist.