Example 3.2.1 Let the absolute values of real numbers $x_{1}, x_{2}, \cdots, x_{n}$ be no greater than 1, find the minimum value of $S=$ $\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}$.
Solution: For any $x_{1}, x_{2}, \cdots, x_{n}, |x_{i}| \leqslant 1$.
For $x_{1}$, let $x_{1}^{\prime}=\left\{\begin{array}{l}-1, x_{1}+x_{2}+\cdots+x_{n}-x_{1} \geqslant 0 ; \\ 1, x_{1}+x_{2}+\cdots+x_{n}-x_{1}<0 .\end{array}\right.$
Adjust $x_{1}, x_{2}, \cdots, x_{n}$ to $x_{1}^{\prime}, x_{2}, \cdots, x_{n}$, the corresponding sum
$S=\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}=x_{1} \cdot \sum_{2 \leqslant i \leqslant n} x_{i}+\sum_{2 \leqslant i<j \leqslant n} x_{i} x_{j} \geqslant x_{1}^{\prime} \sum_{2 \leqslant i \leqslant n} x_{i}+\sum_{2 \leqslant i<j \leqslant n} x_{i} x_{j}=S_{1}$.
Similarly, for $x_{1}^{\prime}, x_{2}, \cdots, x_{n}$, apply the above transformation to $x_{2}$, we also have $S_{1} \geqslant S_{2}, \cdots \cdots$, continuing this process, we get $x_{1}^{\prime} x_{2}^{\prime} \cdots x_{n}^{\prime}$, where $|x_{i}^{\prime}|=1$, and the minimum value of $\sum_{1 \leqslant i \leqslant j \leqslant n} x_{i} x_{j}$ is achieved at $\sum_{1 \leqslant i<j \leqslant n} x_{i}^{\prime} x_{j}^{\prime}$.
$$
S=\sum_{1 \leqslant i<j \leqslant n} x_{i}^{\prime} x_{j}^{\prime}=\frac{1}{2}\left[\left(\sum_{i=1}^{n} x_{i}^{\prime}\right)^{2}-\sum_{i=1}^{n} x_{i}^{\prime 2}\right]=\frac{1}{2}\left[\left(\sum_{i=1}^{n} x_{i}^{\prime}\right)^{2}-n\right] .
$$
When $n$ is even, $S \geqslant-\frac{n}{2}$, and the equality holds when $x_{1}^{\prime} \cdots x_{n}^{\prime}$ contains $\frac{n}{2}$ ones and $\frac{n}{2}$ negative ones,
When $n$ is odd, $S \geqslant-\frac{n-1}{2}$, and the equality holds when $x_{1}^{\prime} \cdots x_{n}^{\prime}$ contains $\frac{n-1}{2}$ negative ones and $\frac{n+1}{2}$ ones, so $\min \sum_{1 \leqslant i \leqslant j \leqslant n} x_{i} y_{j}=-\left[\frac{n}{2}\right]$.