Example 3 Let $x_{i} \geqslant 0, i=1,2, \cdots, n$, and $\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant k<j \leqslant n} \sqrt{\frac{k}{j}} \cdot x_{k} x_{j}=1$.
Find the maximum and minimum values of $\sum_{i=1}^{n} x_{i}$.
(2001, National High School Mathematics Competition)
Explanation: It is easy to see that the minimum value of $\sum_{i=1}^{n} x_{i}$ is 1 (this is achieved when one of the $x_{i}$ is 1 and the rest are all zero). To find the maximum value, note that
$$
\begin{array}{l} 
1=\sum_{i=1}^{n} i\left(\frac{x_{i}}{\sqrt{i}}\right)^{2}+2 \sum_{1 \leq k<j<n} k\left(\frac{x_{k}}{\sqrt{k}} \cdot \frac{x_{j}}{\sqrt{j}}\right) \\
=\sum_{k=1}^{n} k \cdot \frac{x_{k}}{\sqrt{k}}\left(\frac{x_{k}}{\sqrt{k}}+2 \sum_{i=k+1}^{n} \frac{x_{i}}{\sqrt{i}}\right) \\
=\sum_{k=1}^{n} k \cdot \frac{x_{k}}{\sqrt{k}}\left(\sum_{i=k}^{n} \frac{x_{i}}{\sqrt{i}}+\sum_{i=k+1}^{n} \frac{x_{i}}{\sqrt{i}}\right) \\
=\sum_{k=1}^{n} k\left(\sum_{i=k}^{n} \frac{x_{i}}{\sqrt{i}}-\sum_{i=k+1}^{n} \frac{x_{i}}{\sqrt{i}}\right) \cdot \\
\left(\sum_{i=k}^{n} \frac{x_{i}}{\sqrt{i}}+\sum_{i=k+1}^{n} \frac{x_{i}}{\sqrt{i}}\right),
\end{array}
$$

If we let $y_{i}=\sum_{j=i}^{n} \frac{x_{j}}{\sqrt{j}}, i=1,2, \cdots, n$, then all $y_{i} \geqslant 0$. By reversing the proof method of Abel's transformation, the condition can be transformed into $\sum_{i=1}^{n} y_{i}^{2}=1$. From $y_{i}=\sum_{j=i}^{n} \frac{x_{j}}{\sqrt{j}}, i=1,2, \cdots, n$, we have
$$
x_{n}=\sqrt{n} y_{n}, x_{i}=\sqrt{i}\left(y_{i}-y_{i+1}\right), i=1,2 \text {, }
$$
$\cdots, n-1$.
Let $y_{n+1}=0$, then we have
$$
\begin{array}{l}
\sum_{i=1}^{n} x_{i}=\sum_{i=1}^{n} \sqrt{i}\left(y_{i}-y_{i+1}\right) \\
=\sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1}) y_{i} .
\end{array}
$$

Using the Cauchy-Schwarz inequality, the maximum value of $\sum_{i=1}^{n} x_{i}$ is
$$
\sqrt{\sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1})^{2}}
$$
(achieved when $x_{i}=\frac{2 i-\sqrt{i^{2}+i}-\sqrt{i^{2}-i}}{\sqrt{\sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1})^{2}}}, i=1,2, \cdots, n$).