Example 2 Let $n \geqslant 2$ be a positive integer. Find the maximum value of the constant $C(n)$ such that for all real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{i} \in(0,1)(i=1,2$, $\cdot \cdot, n)$, and
$$
\left(1-x_{i}\right)\left(1-x_{j}\right) \geqslant \frac{1}{4}(1 \leqslant i<j \leqslant n)
$$

we have
$$
\sum_{i=1}^{n} x_{i} \geqslant C(n) \sum_{1 \leqslant i<j \leqslant n}\left(2 x_{i} x_{j}+\sqrt{x_{i} x_{j}}\right) .
$$
【Analysis】From the equality of the given inequality, we can set
$$
x_{i}=\frac{1}{2}(i=1,2, \cdots, n) \text {. }
$$

Substituting into equation (1) yields
$$
\frac{n}{2} \geqslant C(n) \mathrm{C}_{n}^{2}\left(\frac{1}{2}+\frac{1}{2}\right) \Rightarrow C(n) \leqslant \frac{1}{n-1} \text {. }
$$

Next, we prove that: $C(n)=\frac{1}{n-1}$ satisfies the condition.
From $1-x_{i}+1-x_{j} \geqslant 2 \sqrt{\left(1-x_{i}\right)\left(1-x_{j}\right)} \geqslant 1$
$$
\Rightarrow x_{i}+x_{j} \leqslant 1 \text {. }
$$

Taking the sum, we get
$$
\begin{array}{l}
(n-1) \sum_{k=1}^{n} x_{k} \leqslant C(n) \Rightarrow \sum_{k=1}^{n} x_{k} \leqslant \frac{n}{2} . \\
\text { Hence } \frac{1}{n-1} \sum_{1 \leqslant i<j \leqslant n}\left(2 x_{i} x_{j}+\sqrt{x_{i} x_{j}}\right) \\
=\frac{2}{n-1}\left(\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}+\sum_{1 \leqslant i<j \leqslant n} \sqrt{x_{i} x_{j}}\right) \\
=\frac{1}{n-1}\left[\left(\sum_{k=1}^{n} x_{k}\right)^{2}-\sum_{k=1}^{n} x_{k}^{2}+\sum_{1 \leqslant i<j<n} \sqrt{x_{i} x_{j}}\right] \\
\leqslant \frac{1}{n-1}\left[\left(\sum_{k=1}^{n} x_{k}\right)^{2}-\frac{1}{n}\left(\sum_{k=1}^{n} x_{k}\right)^{2}+\right. \\
=\frac{1}{n-1}\left[\frac{n-1}{n}\left(\sum_{k=1}^{n} x_{k}\right)^{2}+\frac{n-1}{2} \sum_{k=1}^{n} x_{k}\right] \\
=\frac{1}{n}\left(\sum_{k=1}^{n} x_{k}\right)^{2}+\frac{1}{2} \sum_{k=1}^{n} x_{k} \\
\leqslant \frac{1}{n}\left(\sum_{k=1}^{n} x_{k}\right) \frac{n}{2}+\frac{1}{2} \sum_{k=1}^{n} x_{k}=\sum_{k=1}^{n} x_{k} .
\end{array}
$$

Thus, equation (1) holds.
Therefore, the maximum value of $C(n)$ is $\frac{1}{n-1}$.