Example 6 Let $x_{i} \geqslant 0(i=1,2, \cdots, n)$, 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 $\sum_{i=1}^{n} x_{i}$.
(2001 National High School League Question)
First, find the minimum value.
By the arithmetic mean inequality, we have $\left(\sum_{i=1}^{n} x_{i}\right)^{2}=\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant k<j \leqslant n} x_{k} x_{j} \geqslant 1$, which implies $\sum_{i=1}^{n} x_{i} \geqslant 1$. 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 $\sum_{k=1}^{n} k y_{k}^{2}+2 \sum_{1 \leq k<j \leq n} k y_{k} y_{j}=1$. Let $M=\sum_{k=1}^{n} x_{k}=\sum_{k=1}^{n} \sqrt{k} y_{k}$,

Let $\left\{\begin{array}{l} y_{1}+y_{2}+\cdots+y_{n}=a_{1}, \\ y_{2}+\cdots+y_{n}=a_{2}, \\ \cdots \\ y_{n}=a_{n} .\end{array}\right.$
Then ( *) is equivalent to $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$.
Let $a_{n+1}=0$, then $M=\sum_{k=1}^{n} \sqrt{k}\left(a_{k}-a_{k+1}\right)=\sum_{k=1}^{n} \sqrt{k} a_{k}-\sum_{k=1}^{n} \sqrt{k} a_{k+1}$
$$
=\sum_{k=1}^{n} \sqrt{k} a_{k}-\sum_{k=1}^{n} \sqrt{k-1} a_{k}=\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1}) a_{k} \text {. }
$$

By the Cauchy-Schwarz inequality, we get
$$
M \leqslant\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}} \cdot\left(\sum_{k=1}^{n} a_{k}^{2}\right)^{\frac{1}{2}}=\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}} \text {. }
$$

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

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

Therefore, the maximum value is $\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}$.