14. Let $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}_{+}$, define
$$
S_{n}=\sum_{i=1}^{n}\left(x_{i}+\frac{n-1}{n^{2}} \cdot \frac{1}{x_{i}}\right)^{2} \text {. }
$$
(1) Find the minimum value of $S_{n}$;
(2) Under the condition $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$, find the minimum value of $S_{n}$;
(3) Under the condition $x_{1}+x_{2}+\cdots+x_{n}=1$, find the minimum value of $S_{n}$, and provide a proof.
14. (1) $S_{n} \geqslant \sum_{i=1}^{n}\left(2 \sqrt{\frac{n-1}{n^{2}}}\right)^{2}$
$$
=4 \sum_{i=1}^{n} \frac{n-1}{n^{2}}=4 \times \frac{n-1}{n} \text {. }
$$

When $x_{1}=x_{2}=\cdots=x_{n}=\frac{\sqrt{n-1}}{n}$, the minimum value $4 \times \frac{n-1}{n}$ is achieved.
$$
\begin{array}{l}
\text { (2) } S_{n}=\sum_{i=1}^{n}\left(x_{i}^{2}+2 \times \frac{n-1}{n^{2}}+\frac{(n-1)^{2}}{n^{4}} \cdot \frac{1}{x_{i}^{2}}\right) \\
=1+2 \times \frac{n-1}{n}+\frac{(n-1)^{2}}{n^{4}} \sum_{i=1}^{n} \frac{1}{x_{i}^{2}} \\
\geqslant 1+2 \times \frac{n-1}{n}+\frac{(n-1)^{2}}{n^{2}}=\left(1+\frac{n-1}{n}\right)^{2} .
\end{array}
$$

When $x_{1}=x_{2}=\cdots=x_{n}=\frac{1}{\sqrt{n}}$, the minimum value
$$
\left(1+\frac{n-1}{n}\right)^{2}
$$
is achieved.
(3) Since $\left[\sum_{i=1}^{n} 1 \times\left(x_{i}+\frac{n-1}{n^{2}} \cdot \frac{1}{x_{i}}\right)\right]^{2}$
$$
\leqslant\left(\sum_{i=1}^{n} 1^{2}\right) \cdot \sum_{i=1}^{n}\left(x_{i}+\frac{n-1}{n^{2}} \cdot \frac{1}{x_{i}}\right)^{2} \text {, }
$$

Therefore, $S_{n}=\sum_{i=1}^{n}\left(x_{i}+\frac{n-1}{n^{2}} \cdot \frac{1}{x_{i}}\right)^{2}$
$$
\begin{array}{l}
\geqslant \frac{1}{n}\left[\sum_{i=1}^{n}\left(x_{i}+\frac{n-1}{n^{2}} \cdot \frac{1}{x_{i}}\right)\right]^{2} \\
\geqslant \frac{1}{n}\left[1+\frac{n-1}{n^{2}} \cdot n^{2}\right]^{2}=n .
\end{array}
$$

When $x_{1}=x_{2}=\cdots=x_{n}=\frac{1}{n}$, the minimum value $n$ is achieved.