2. 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 $\sum_{i=1}^{n} x_{i}$.
Since $\left(\sum_{i=1}^{n} x_{i}\right)^{2}=\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant k<j \leqslant n} x_{k} x_{j} \geqslant 1$, equality holds if and only if there exists $i$ such that $x_{i}=1$ and the rest $x_{j}=0, j \neq i$, so the minimum value of $\sum_{i=1}^{n} x_{i}$ is 1.
Let $x_{k}=\sqrt{k} y_{k}(1 \leqslant k \leqslant n)$, then $\sum_{k=1}^{n} k y_{k}^{2}+2 \sum_{1 \leqslant k<j \leqslant n} k y_{k} y_{j}=1$.
Let $M=\sum_{k=1}^{n} x_{k}=\sum_{k=1}^{n} \sqrt{k} y_{k}$, and let $\left\{\begin{array}{l}y_{1}+y_{2}+\cdots+y_{n}=a_{1}, \\ y_{2}+y_{3}+\cdots+y_{n}=a_{2}, \\ \cdots \cdots \\ y_{n}=a_{n},\end{array}\right.$
then (1) is equivalent to $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$. Let $a_{n+1}=0$,
thus $M=\sum_{k=1}^{n} \sqrt{k}\left(a_{k}-a_{k+1}\right)=\sum_{k=1}^{n} \sqrt{k} a_{k}-\sum_{k=1}^{n} \sqrt{k} a_{k+1}=\sum_{k=1}^{n} \sqrt{k} a_{k}-\sum_{k=1}^{n} \sqrt{k-1} a_{k}$
$=\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1}) a_{k} \leqslant\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}} \cdot\left(\sum_{k=1}^{n} a_{k}^{2}\right)^{\frac{1}{2}}$
$=\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}$, equality holds when $\frac{a_{1}^{2}}{1}=\frac{a_{2}^{2}}{(\sqrt{2}-1)^{2}}=\cdots=\frac{a_{n}^{2}}{(\sqrt{n}-\sqrt{n-1})^{2}}$
$\Leftrightarrow \frac{a_{k}^{2}}{(\sqrt{k}-\sqrt{k-1})^{2}}=\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}{1+(\sqrt{2}-1)^{2}+\cdots+(\sqrt{n}-\sqrt{n-1})^{2}}$
$\Leftrightarrow a_{k}=\frac{\sqrt{k}-\sqrt{k-1}}{\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}}(k=1,2,3, \cdots, n)$. Since $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}$
$\Rightarrow y_{k}=a_{k}-a_{k+1}=\frac{2 \sqrt{k}-(\sqrt{k+1}+\sqrt{k-1})}{\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}} \geqslant 0 \Rightarrow x_{k} \geqslant 0$,
so the maximum value of $\sum_{i=1}^{n} x_{i}$ is $\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}$.