Let $x_{1}, x_{2}, \ldots, x_{n}$ be numbers whose absolute values are at most 1. What is the smallest possible value of

$$
x_{1} x_{2}+x_{1} x_{3}+\cdots+x_{1} x_{n}+x_{2} x_{3}+\cdots+x_{n-1} x_{n}
$$
Consider an arbitrary $n$-tuple of numbers $x_{1}, x_{2}, \ldots, x_{n}$, where $-1 \leq x_{i} \leq 1$ for $i=1,2, \ldots, n$. Let the expression $x_{1} x_{2}+x_{1} x_{3}+\ldots x_{n-1} x_{n}$ be denoted by $S$. We will show that the value of $S$ does not increase if the $x_{i}$ are appropriately made to have an absolute value of 1. Collecting the terms involving $x_{1}$,

$$
S=x_{1}\left(x_{2}+x_{3}+\cdots+x_{n}\right)+x_{2} x_{3}+\cdots+x_{n-1} x_{n}
$$

if $x_{2}+x_{3}+\cdots+x_{n} \geq 0$, then by choosing $x_{1}^{*}=-1$, the resulting sum $S^{*}$ will satisfy $S^{*} \leq S$, while if $x_{2}+x_{3}+\cdots+x_{n}<0$, then by choosing $x_{1}^{*}=+1$, we will have $S^{*} \leq S$. We then perform the same process sequentially for $x_{2}$, $x_{3}$, $\ldots, x_{n}$.

Based on these, when searching for the minimum, we can assume that $\left|x_{i}\right|=1$ for $i=1,2, \ldots, n$, since $S$ will certainly achieve its minimum for such $n$-tuples. Write $S$ in the following form:

$$
S=\frac{\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2}-\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)}{2}
$$

Here, $\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)=n$, so

$$
S=\frac{\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2}}{2}-\frac{n}{2} \geq-\frac{n}{2}
$$

equality holds precisely when $x_{1}+x_{2}+\cdots+x_{n}=0$. This is achievable for even $n$, where $n / 2$ of the $x_{i}$ are +1 and the rest are -1. For odd $n$, $\left|x_{1}+x_{2}+\cdots+x_{n}\right| \geq 1$, thus

$$
S \geq \frac{1}{2}-\frac{n}{2}=\frac{1-n}{2}
$$

and this is achievable: take one more +1 than -1.

The sought minimum value is $-\frac{n}{2}$ for even $n$, and $\frac{1-n}{2}$ for odd $n$.

Based on the work of Fejes Tóth Péter (Budapest, Árpád Gym., II. o. t.)