Three. (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 $\sum_{i=1}^{n} x_{i}$.
Three, Solution: First, find the minimum value, because $\left(\sum_{i=1}^{n} x_{i}\right)^{2}=\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1<k<j<n} \sqrt{\frac{k}{j}} x_{k} x_{j} \geqslant 1-\Rightarrow \sum_{i=1}^{n} x_{i} \geqslant 1$, equality holds if and only if there exists $i$ such that $x_{i}=1, x_{j}=0, j=i$.
$\therefore \sum_{i=1}^{n} x_{i}$ minimum value is 1.
Next, find the maximum value, let $x_{k}=\sqrt{k} y_{k}$,
$$
\therefore \sum_{k=1}^{n} k y_{k}^{2}+2 \sum_{1<k<j<n} k y_{k} y_{j}=1
$$

Let $M=\sum_{k=1}^{n} x_{k}=\sum_{k=1}^{n} \sqrt{k} y_{k}$, and let $\left\{\begin{array}{l}y_{1}+y_{2}+\cdots+y_{n}=a_{1}, \\ y_{2}+\cdots+y_{n}=a_{2}, \\ \cdots \cdots \\ y_{n}=a_{n} .\end{array}\right.$
Then (1) $\Leftrightarrow a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$, let $a_{n-1}=0$, then $M=\sum_{k=1}^{n} \sqrt{k}\left(a_{k}-a_{k+1}\right)$
$$
=\sum_{k=1}^{n} \sqrt{k} a_{k}-\sum_{k=1}^{n} \sqrt{k} a_{k+1}=\sum_{k=1}^{n} \sqrt{k} a_{k}-\sum_{k=1}^{n} \sqrt{k-1} a_{k}=\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1}) a_{k} .
$$

By the Cauchy-Schwarz inequality, we get: $M \leqslant\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}\left(\sum_{k=1}^{n} a_{k}^{2}\right)^{\frac{1}{2}}=\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}$,
equality holds $\Leftrightarrow \frac{a_{1}^{2}}{1}=\cdots=\frac{a_{k}^{2}}{(\sqrt{k}-\sqrt{k-1})^{2}}=\cdots=\frac{a_{n}^{2}}{(\sqrt{n}-\sqrt{n-1})^{2}}$
$\Leftrightarrow \frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}{1+(\sqrt{2}-\sqrt{1})^{2}+\cdots+(\sqrt{n}-\sqrt{n-1})^{2}}=\frac{a_{k}^{2}}{(\sqrt{k}-\sqrt{k-1})^{2}}$
$\Leftrightarrow a_{k}=\frac{\sqrt{k}-\sqrt{k-1}}{\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}} \quad(k=1,2, \cdots, n)$, 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 $\left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}}$.