II. (This question is worth 50 points)
Let $x_{i} \geqslant 0(i=1,2,3, \cdots, n)$ and $\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant k<j \leqslant n} \sqrt{\frac{k}{j}} x_{k} x_{j}=1$,
find the maximum and minimum values of $\sum_{i=1}^{n} x_{i}$.
$$
\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}, 1
$$

II. 【Analysis and Solution】First, find the minimum value. Because $\left(\sum_{i=1}^{n} x_{i}\right)^{2}=\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant k<j \leqslant n} \sqrt{\frac{k}{j}} x_{k} x_{j} \geqslant 1 \Rightarrow \sum_{i=1}^{n} x_{i} \geqslant 1$, equality holds if and only if there exists $i$ such that $x_{i}=1, x_{j}=0, j \neq i$.
$\therefore \sum_{i=1}^{n} x_{i}$ has a minimum value of 1. $\qquad$ 10 points
Next, find the maximum value. Let $x_{k}=\sqrt{k} y_{k}$
$\therefore \sum_{k=1}^{n} k y_{k}^{2}+2 \sum_{1 \leqslant k<j \leqslant n} k y_{k} y_{j}=1$
Let $M=\sum_{k=1}^{n} x_{k}=\sum_{k=1}^{n} \sqrt{k} y_{k}$, and let $\begin{array}{l} y_{1}+y_{2}+\cdots+y_{n}=a_{1} \\ y_{2}+\cdots+y_{n}=a_{2} \\ \cdots \\ y_{n}=a_{n}\end{array}$
(1) $\Leftrightarrow a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$. $\qquad$ 30 points
Let $a_{n-1}=0$, then $M=\sum_{k=1}^{n} \sqrt{k}\left(a_{k}-a_{k+1}\right)$
$=\sum_{k=1}^{n} \sqrt{k} a_{k}-\sum_{k=1}^{n} \sqrt{k} a_{k+1}=\sum_{k=1}^{n} \sqrt{k} a_{k}-\sum_{k=1}^{n} \sqrt{k-1} a_{k}=\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1}) a_{k}$.
By the Cauchy-Schwarz inequality:
$M \leqslant\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}\left(\sum_{k=1}^{n} a_{k}^{2}\right)^{\frac{1}{2}}=\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}$,
Equality holds $\Leftrightarrow \frac{a_{1}^{2}}{1}=\cdots=\frac{a_{k}^{2}}{(\sqrt{k}-\sqrt{k-1})^{2}}=\cdots=\frac{a_{n}^{2}}{(\sqrt{n}-\sqrt{n-1})^{2}}$
$\Leftrightarrow \frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}{1+(\sqrt{2}-\sqrt{1})^{2}+\cdots+(\sqrt{n}-\sqrt{n-1})^{2}}=\frac{a_{k}^{2}}{(\sqrt{k}-\sqrt{k-1})^{2}}$ $\Leftrightarrow a_{k}=\frac{\sqrt{k}-\sqrt{k-1}}{\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}}(k=1,2, \cdots, n)$ Since $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}$, thus $y_{k}=a_{k}-a_{k+1}=\frac{2 \sqrt{k}-(\sqrt{k+1}+\sqrt{k-1})}{\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}} \geqslant 0$,
i.e., $x_{k} \geqslant 0$,
The maximum value sought is $\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}$. $\qquad$