$x_{2}+\cdots+x_{n}=1$, let
$$
S=\max \left\{\frac{x_{1}}{1+x_{1}}, \frac{x_{2}}{1+x_{1}+x_{2}}, \cdots, \frac{x_{n}}{1+x_{1}+\cdots+x_{n}}\right\},
$$

find the minimum value of $S$.
Let $y_{0}=1, y_{k}=1+x_{1}+\cdots+x_{k}, 1 \leqslant k \leqslant n$, then $y_{n}=2, x_{k}=y_{k}-y_{k-1}, k=1,2, \cdots, n$. Thus, $S=\max \left\{\frac{x_{1}}{y_{1}}, \frac{x_{2}}{y_{2}}, \cdots, \frac{x_{n}}{y_{n}}\right\}$.
Since $\frac{x_{k}}{y_{k}} \leqslant S, k=1,2, \cdots, n$, we have
$$
\begin{array}{l}
S \geqslant \frac{x_{k}}{y_{k}}=\frac{y_{k}-y_{k-1}}{y_{k}}=1-\frac{y_{k-1}}{y_{k}}, 1-S \leqslant \frac{y_{k-1}}{y_{k}}, \\
(1-S)^{n} \leqslant \frac{y_{0}}{y_{1}} \cdot \frac{y_{1}}{y_{2}} \cdots \cdot \frac{y_{n-1}}{y_{n}}=\frac{y_{0}}{y_{n}}=\frac{1}{2}, S \geqslant 1-\frac{1}{2^{\frac{1}{n}}} .
\end{array}
$$

When $y_{1}=2^{\frac{1}{n}}, y_{2}=2^{\frac{2}{n}}, \cdots, y_{n}=2$, i.e., $x_{k}=2^{\frac{k}{n}}-2^{\frac{k-1}{n}}$, the equality holds.
Therefore, the minimum value of $S$ is $1-\frac{1}{2^{\frac{1}{n}}}$.