2. Let $x_{i} \in \mathbf{R},\left|x_{i}\right| \leqslant 1(1 \leqslant i \leqslant n)$, find the minimum value of $F=\sum_{1 \leq i<j \leqslant n} x_{i} x_{j}$.
2. Since $F$ is continuous on a closed domain, there must exist a minimum value. Fixing $x_{2}, x_{3}, \cdots, x_{n}$, then $F\left(x_{1}\right)$ is a linear function of $x_{1}$. Given $-1 \leqslant x_{1} \leqslant 1$, when $F\left(x_{1}\right)$ takes its minimum value, it must be that $x_{1} \in\{-1,1\}$. By symmetry, when $F$ takes its minimum value, it must be that $x_{i} \in\{-1,1\}(1 \leqslant i \leqslant n)$. Suppose that when $F$ takes its minimum value, there are $k$ values of $x_{i}$ that are $1$ and $n-k$ values that are $-1$, then the minimum value of $F$ is the minimum value of $F(k)$ in the following form: $F(k)=\mathrm{C}_{k}^{2}+\mathrm{C}_{n-k}^{2}-\mathrm{C}_{k}^{1} \mathrm{C}_{n-k}^{1}=2\left(k-\frac{n}{2}\right)^{2}-\frac{n}{2} \geqslant-\frac{n}{2}$. But $F(k)$ is an integer, so $F(k) \geqslant-\left[\frac{n}{2}\right]$, where equality holds when $k=\left[\frac{n}{2}\right]$. Therefore, the minimum value of $F$ is $-\left[\frac{n}{2}\right]$. Alternatively, when $x_{i} \in\{-1,1\}$, $F=\frac{1}{2}\left(\sum_{i=1}^{n} x_{i}\right)^{2}-\sum_{i=1}^{n} \frac{1}{2} x_{i}^{2}=$ $\frac{1}{2}\left(\sum_{i=1}^{n} x_{i}\right)^{2}-\sum_{i=1}^{n} \frac{1}{2}=\frac{1}{2}\left(\sum_{i=1}^{n} x_{i}\right)^{2}-\frac{n}{2} \geqslant-\frac{n}{2}$, thus $-F \leqslant \frac{n}{2}$. But $-F$ is an integer, so $-F \leqslant\left[\frac{n}{2}\right]$, which means $F \geqslant-\left[\frac{n}{2}\right]$.