$$
\begin{array}{l}
\quad \text { II. (50 points) } \\
\quad \text { Let } x_{i} \geqslant 0(i=1,2, \cdots, n), \text { and } \sum_{i=1}^{n} x_{i}^{2}+ \\
2 \sum_{1 \leq k < j \leq n} x_{k} x_{j}=1 .
\end{array}
$$

Find the maximum and minimum values of $\sum_{i=1}^{n} x_{i}$.

II. (50 points)

First, find the minimum value.
Since $\left(\sum_{i=1}^{n} x_{i}\right)^{2}-\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant k<j \leqslant n} x_{k} x_{j} \geqslant 1 \Rightarrow \sum_{i=1}^{n} x_{i} \geqslant 1$,
the equality holds if and only if there exists $i$ such that $x_{i}=1, x_{j}=0, j \neq i$.
Therefore, the minimum value of $\sum_{i=1}^{n} x_{i}$ is 1.
Next, find the maximum value. Let $x_{k}=\sqrt{k} y_{k}$. Then,
$$
\sum_{k=1}^{n} k y_{k}^{2}+2 \sum_{1 \leqslant k < j \leqslant n} k y_{k} y_{j}=1 .
$$

Let $M=\sum_{k=1}^{n} x_{k}=\sum_{k=1}^{n} \sqrt{k} y_{k}$.
Let $\left\{\begin{array}{l} y_{1}+y_{2}+\cdots+y_{n}=a_{1}, \\ y_{2}+\cdots+y_{n}=a_{2}, \\ \cdots \cdots \\ y_{n}=a_{n} .\end{array}\right.$
Then (1) $\Leftrightarrow a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$.
Let $a_{n+1}=0$. Then
$$
\begin{aligned}
M & =\sum_{k=1}^{n} \sqrt{k}\left(a_{k}-a_{k+1}\right)=\sum_{k=1}^{n} \sqrt{k} a_{k}-\sum_{k=1}^{n} \sqrt{k} a_{k+1} \\
& =\sum_{k=1}^{n} \sqrt{k} a_{k}-\sum_{k=1}^{n} \sqrt{k-1} a_{k}=\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1}) a_{k} .
\end{aligned}
$$

By the Cauchy-Schwarz inequality,
$$
\begin{array}{c}
M \leqslant\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}\left(\sum_{k=1}^{n} a_{k}^{2}\right)^{\frac{1}{2}} \\
=\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}
\end{array}
$$
The equality holds $\Leftrightarrow \frac{a_{1}^{2}}{1}=\cdots=\frac{a_{k}^{2}}{(\sqrt{k}-\sqrt{k-1})^{2}}=\cdots=\frac{a_{n}^{2}}{(\sqrt{n}-\sqrt{n-1})^{2}}$
$$
\begin{array}{l}
\Leftrightarrow \frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}{1+(\sqrt{2}-\sqrt{1})^{2}+\cdots+(\sqrt{n}-\sqrt{n-1})^{2}}=\frac{a_{k}^{2}}{(\sqrt{k}-\sqrt{k-1})^{2}} \\
\Leftrightarrow a_{k}=\frac{\sqrt{k}-\sqrt{k-1}}{\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}} \cdot(k=1,2, \cdots, n)
\end{array}
$$

Since $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}$, thus,
$$
y_{k}=a_{k}-a_{k+1}=\frac{2 \sqrt{k}-(\sqrt{k+1}+\sqrt{k-1})}{\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}} \geq 0,
$$

which means $x_{k} \geqslant 0$.
The maximum value sought is $\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}$.