Given a positive integer $n(n \geqslant 2)$. Find the largest positive real number $C$, such that for any $x_{1}, x_{2}, \cdots, x_{n}$, we have
$$
\sum_{i=1}^{n} x_{i}^{2} \geqslant \frac{1}{n+1}\left(\sum_{i=1}^{n} x_{i}\right)^{2}+C\left(\sum_{i=1}^{n} i x_{i}\right)^{2} .
$$
$$
\text { Two, } C_{\max }=\frac{12}{n(n+1)(n+2)(3 n+1)} \text {. }
$$

Let $S_{i}=\sum_{j=i}^{n} x_{j}(1 \leqslant i \leqslant n)$. Then
$$
\begin{array}{l}
S_{i}-S_{i+1}=x_{i}(1 \leqslant i \leqslant n-1) . \\
\text { Hence } \sum_{i=1}^{n} S_{i}=\sum_{i=1}^{n} \sum_{j=i}^{n} x_{j}=\sum_{j=1}^{n} j x_{j},
\end{array}
$$

and.
$$
\begin{array}{l}
\sum_{i=1}^{n} x_{i}^{2}-\frac{1}{n+1}\left(\sum_{i=1}^{n} x_{i}\right)^{2} \\
=-\frac{S_{1}^{2}}{n+1}+\sum_{i=1}^{n-1}\left(S_{i}-S_{i+1}\right)^{2}+S_{n}^{2} \\
=\frac{n}{n+1} S_{1}^{2}+2 \sum_{i=2}^{n} S_{i}^{2}-2 \sum_{i=1}^{n-1} S_{i} S_{i+1} \\
=\frac{S_{1}^{2}}{n(n+1)}+ \\
=\frac{\sum_{i=1}^{n-1}\left(\frac{i+1}{i} S_{n+1-i}^{2}-2 S_{n+1-i} S_{n-i}+\frac{i}{i+1} S_{n-i}^{2}\right)}{n(n+1)}+\sum_{i=1}^{n-1} \frac{\left((i+1) S_{n+1-i}-i S_{n-i}\right)^{2}}{i(i+1)} \\
\geqslant \frac{\left((n+1) S_{1}+\sum_{i=1}^{n-1}\left((i+1) S_{n+1-i}-i S_{n-i}\right)\right)^{2}}{n(n+1)^{3}+\sum_{i=1}^{n-1} i(i+1)} \\
=\frac{3\left(2 \sum_{i=1}^{n} S_{i}\right)^{2}}{n(n+1)(n+2)(3 n+1)} .
\end{array}
$$

Combining with equation (1), we have
$$
\begin{array}{l}
\frac{\sum_{i=1}^{n} x_{i}^{2}-\frac{1}{n+1}\left(\sum_{i=1}^{n} x_{i}\right)^{2}}{\left(\sum_{i=1}^{n} i x_{i}\right)^{2}} \\
\geqslant \frac{12}{n(n+1)(n+2)(3 n+1)}
\end{array}
$$

Thus, $C \leqslant \frac{12}{n(n+1)(n+2)(3 n+1)}$.
Furthermore, when the equality in the above inequality holds, i.e., $x_{i}=n^{2}+$ $n+2 i(1 \leqslant i \leqslant n)$, we have
$$
\begin{array}{l}
\sum_{i=1}^{n} x_{i}^{2}-\frac{1}{n+1}\left(\sum_{i=1}^{n} x_{i}\right)^{2} \\
=\frac{12}{n(n+1)(n+2)(3 n+1)}\left(\sum_{i=1}^{n} i x_{i}\right)^{2}
\end{array}
$$

Therefore, $C_{\max }=\frac{12}{n(n+1)(n+2)(3 n+1)}$.