$\square$ Example 17 Let $x_{i} \geqslant 0(i=1,2, \cdots, n)$ and $\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant i<j \leqslant n} \sqrt{\frac{i}{j}} x_{i} x_{j}$ $=1$, find the maximum and minimum values of $\sum_{i=1}^{n} x_{i}$. (2001 National High School Mathematics League Additional Question)
Solve for the minimum value first. Since
$$\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j} \geqslant \sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant i<j \leqslant n} \sqrt{\frac{i}{j}} x_{i} x_{j}=1$$

Therefore, $\sum_{i=1}^{n} x_{i} \geqslant 1$, equality holds if and only if there exists an $i$ such that $x_{i}=1, x_{j}=0(j \neq i)$, so the minimum value of $\sum_{i=1}^{n} x_{i}$ is 1.

Next, solve for the maximum value. Let $x_{k}=\sqrt{k} y_{k}(k=1,2, \cdots, n)$, so
$$\sum_{k=1}^{n} k y_{k}^{2}+2 \sum_{1 \leqslant k<j \leqslant n} k y_{k} y_{j}=1$$

Let $M=\sum_{k=1}^{n} x_{k}=\sum_{k=1}^{n} \sqrt{k} y_{k}$, and
$$\left\{\begin{array}{l}
y_{1}+y_{2}+\cdots+y_{n}=a_{1} \\
y_{2}+\cdots+y_{n}=a_{2} \\
\cdots \quad \cdots \quad \cdots \\
y_{n}=a_{n}
\end{array}\right.$$

Then (1) is equivalent to
$$a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$$

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,
$$\begin{aligned}
M & \leqslant\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}\left(\sum_{k=1}^{n} a_{k}^{2}\right)^{\frac{1}{2}} \\
& =\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}
\end{aligned}$$

Equality holds $\Leftrightarrow \frac{a_{1}^{2}}{1}=\frac{a_{2}^{2}}{(\sqrt{2}-1)^{2}}=\cdots=\frac{a_{k}^{2}}{(\sqrt{k}-\sqrt{k-1})^{2}}=\cdots$
$$\begin{aligned}
& \Rightarrow \frac{a_{n}^{2}}{(\sqrt{n}-\sqrt{n-1})^{2}} \\
\Leftrightarrow & \frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}{1+(\sqrt{2}-1)^{2}+\cdots+(\sqrt{n}-\sqrt{n-1})^{2}} \\
& =\frac{a_{k}^{2}}{(\sqrt{k}-\sqrt{k-1})^{2}} \\
\Leftrightarrow & a_{k}=\frac{\sqrt{k}-\sqrt{k-1}}{\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}}(k=1,2, \cdots, n) .
\end{aligned}$$

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 $\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}$.