[Proof] For any permutation $\pi=\left(y_{1}, y_{2}, \cdots, y_{n}\right)$ of $x_{1}, x_{2}, \cdots, x_{n}$. Let $S(\pi)=y_{1}+2 y_{2}+\cdots+n y_{n}$.

Let $r=\frac{n+1}{2}$.
We need to prove that there exists a permutation $\pi$ such that $|S(\pi)| \leqslant r$.
$$\text { Let } \pi_{0}=\left(x_{1}, x_{2}, \cdots, x_{n}\right), \bar{\pi}=\left(x_{n}, x_{n-1}, \cdots, x_{1}\right) .$$

If $\left|S\left(\pi_{0}\right)\right| \leqslant r$ or $|S(\pi)| \leqslant r$, then the problem is solved. If $\left|S\left(\pi_{0}\right)\right|>r$ and $|S(\bar{\pi})|>r$. Note that
$$\begin{aligned} S\left(\pi_{0}\right)+S(\bar{\pi}) & =\left(x_{1}+2 x_{2}+\cdots+n x_{n}\right)+\left(x_{n}+2 x_{n-1}+\cdots+n x_{1}\right) \\ & =(n+1)\left(x_{1}+x_{2}+\cdots+x_{n}\right), \end{aligned}$$

so $\quad\left|S\left(\pi_{0}\right)+S(\bar{\pi})\right|=n+1=2 r$.
Since $\mid S\left(\pi_{0}\right)$ | and $|S(\bar{\pi})|$ are both greater than $r$, $S\left(\pi_{0}\right)$ and $S(\bar{\pi})$ have opposite signs, and one of them is greater than $r$, the other is less than $-r$.

Starting from $\pi_{0}$, by swapping the positions of two adjacent elements several times, we can obtain any permutation. In particular, there exists a sequence of permutations
$$\pi_{0}, \pi_{1}, \pi_{2}, \cdots, \pi_{m},$$

where $\pi_{m}=\bar{\pi}$, and for each $i \in\{0,1, \cdots, m-1\}$, permutation $\pi_{i+1}$ is obtained by swapping the positions of two adjacent elements in $\pi_{i}$. That is, if
$\pi_{i}=\left(y_{1}, y_{2}, \cdots, y_{n}\right), \pi_{i+1}=\left(z_{1}, z_{2}, \cdots, z_{n}\right)$, then there exists $k \in\{1,2$, $\cdots, n-1\}$, such that
$z_{k}=y_{k+1}, z_{k+1}=y_{k} ; z_{j}=y_{j}, j \notin\{k, k+1\}$ when. Since $\left|x_{i}\right| \leqslant r$, $i=1,2, \cdots, n$, we have
$$\begin{aligned} & \left|S\left(\pi_{i+1}\right)-S\left(\pi_{i}\right)\right|=\left|k z_{k}+(k+1) z_{k+1}-k y_{k}-(k+1) y_{k+1}\right| \\ = & \left|y_{k}-y_{k+1}\right| \leqslant\left|y_{k}\right|+\left|y_{k+1}\right| \leqslant 2 r . \end{aligned}$$

This shows that in the sequence $S\left(\pi_{0}\right), S\left(\pi_{1}\right), \cdots, S\left(\pi_{m}\right)$, the distance between any two adjacent numbers does not exceed $2 r$. Noting that $S\left(\pi_{0}\right)$ and $S\left(\pi_{m}\right)$ both fall outside the interval $[-r, r]$, and are on opposite sides of the interval, at least one number $S\left(\pi_{i}\right)$ falls within the interval. That is, there exists a permutation $\pi_{i}$ such that
$$\left|S\left(\pi_{i}\right)\right| \leqslant r .$$