Example 1.19 (Han Jingjun) $x_{i} \geqslant 0, i=1,2, \cdots, n . \sum_{i=1}^{n} x_{i}=1$, find
$$\max \left\{x_{1}+x_{2}+\cdots+x_{j}, x_{2}+x_{3}+\cdots+x_{j+1}, \cdots, x_{n-j+1}+x_{n-j+2} \cdots+x_{n}\right\}$$

the minimum value.
Notice
$$\begin{array}{l}
\max \left\{x_{1}+x_{2}+\cdots+x_{j}, x_{2}+x_{3}+\cdots+x_{j+1}, \cdots, x_{n-j+1}+x_{n-j+2}+\cdots+x_{n}\right\} \geqslant \\
\max \left\{x_{n-j+1}+x_{n-j+2}+\cdots+x_{n}, x_{1}+x_{2}+\cdots+x_{j}, x_{j+1}+x_{j+2}+\cdots+\right. \\
\left.x_{2 j}, \cdots, x_{\left[\frac{n}{j}\right]+1-j}+x_{\left[\frac{n}{j}\right]+2-j}+\cdots+x_{\left[\frac{n}{j}\right] j}\right\}
\end{array}$$

When $j x_{n}$, let $n=j k+l, j-1 \geqslant l \geqslant 1$. The minimum value $\geqslant \frac{1}{\left[\frac{n}{j}\right]+1}=\frac{1}{\left[\frac{n+j-1}{j}\right]}$. At this time, set $x_{i}=0, i \equiv l+1, l+2, \cdots, j(\bmod j)$, and the rest of $x_{i}$ are equal.
When $j \backslash n$, the minimum value $\geqslant \frac{1}{\left[\frac{n}{j}\right]}=\frac{1}{\left[\frac{n+j-1}{j}\right]}$. At this time, set all $x_{j}$ equal. In summary, the minimum value is $\frac{1}{\left[\frac{n+j-1}{j}\right]}$.