When $x_{0}=x_{1}=\cdots=x_{n}=\frac{1}{n+1}$,
$$
S=\sqrt{n+1} \text {. }
$$

Next, we prove that this is the maximum value.
Notice,
$$
\begin{array}{l}
\left(\sqrt{x_{i}}+\sqrt{x_{i+1}}\right)^{2} \leqslant 2\left(x_{i}+x_{i+1}\right) \leqslant 2 . \\
\text { Then } \frac{\left(x_{i}-x_{i+1}\right)^{2}}{2 n} \leqslant \frac{\left(\sqrt{x_{i}}-\sqrt{x_{i+1}}\right)^{2}}{n} \\
\Rightarrow \sqrt{x_{0}+\sum_{i=1}^{n} \frac{\left(x_{i}-x_{i+1}\right)^{2}}{2 n}} \\
\quad \leqslant \sqrt{x_{0}+\sum_{i=1}^{n} \frac{\left(\sqrt{x_{i}}-\sqrt{x_{i+1}}\right)^{2}}{n}} \\
\quad=\sqrt{1-\left(\left(1-\frac{2}{n}\right) \sum_{i=1}^{n} x_{i}+\frac{2}{n} \sum_{i=1}^{n} \sqrt{x_{i} x_{i+1}}\right)} .
\end{array}
$$

We now prove:
$$
\begin{array}{l}
\left(1-\frac{2}{n}\right) \sum_{i=1}^{n} x_{i}+\frac{2}{n} \sum_{i=1}^{n} \sqrt{x_{i} x_{i+1}} \\
\geqslant \frac{1}{n}\left(\sum_{i=1}^{n} \sqrt{x_{i}}\right)^{2} .
\end{array}
$$

In fact, by the mean inequality,
$$
\begin{array}{l}
\left(\sum_{i=1}^{n} \sqrt{x_{i}}\right)^{2}-2 \sum_{i=1}^{n} \sqrt{x_{i} x_{i+1}} \\
=\sum_{i=1}^{n} x_{i}+\sum_{\substack{3 \leqslant j+2 \leqslant i \leqslant n \\
(i, j) \neq(1, n)}} 2 \sqrt{x_{i} x_{j}} \\
\leqslant \sum_{i=1}^{n} x_{i}+\sum_{\substack{3 \leqslant j+2 \leqslant i \leqslant n \\
(i, j) \neq(1, n) \\
n)}}\left(x_{i}+x_{j}\right) \\
=(n-2) \sum_{i=1}^{n} x_{i},
\end{array}
$$

which is clearly true.
Let $t=\sum_{i=1}^{n} \sqrt{x_{i}}$.
Then by the Cauchy-Schwarz inequality,
$$
\begin{array}{l}
S \leqslant \sqrt{1-\frac{t^{2}}{n}}+t=\sqrt{1-\frac{t^{2}}{n}} \cdot 1+\frac{t}{\sqrt{n}} \cdot \sqrt{n} \\
\leqslant \sqrt{1} \cdot \sqrt{1+n}=\sqrt{n+1} .
\end{array}
$$

In summary, the maximum value of $S$ is $\sqrt{n+1}$.
One, (40 points) For non-negative real numbers $x_{0}$, $x_{1}, \cdots, x_{n}$ that sum to 1, and with the convention that $x_{n+1}=x_{1}$, find
$$
S=\sqrt{x_{0}+\sum_{i=1}^{n} \frac{\left(x_{i}-x_{i+1}\right)^{2}}{2 n}}+\sum_{i=1}^{n} \sqrt{x_{i}}(n \geqslant 3)
$$

the maximum value.