Two. (50 points) Given $x_{i} \geqslant 0(i=1,2, \cdots, n), n \geqslant 2$, and $\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant k<j \leqslant n} \frac{k}{j} x_{k} x_{j}=1$. Try to find the maximum and minimum values of $\sum_{j=1}^{n} x_{i}$.
$$
\begin{array}{l}
\because\left(\sum_{i=1}^{n} x_{i}\right)^{2} \geqslant \sum_{i=1}^{n} x_{i}^{2}+2 \sum_{i \leqslant i<i \leqslant n} \frac{k}{j} x_{i} x_{j}=1, \\
\therefore \sum_{i=1}^{n} x_{i} \geqslant 1 .
\end{array}
$$

Therefore, the minimum value of $\sum_{i=1}^{n} x_{i}$ is 1, and the equality holds if and only if there exists $i$ such that $x_{i}=1, x_{j}=0(j \neq i)$.
Next, let's find the maximum value.
Let $x_{k}=k y_{i}, 1 \leqslant k \leqslant n$. The original expression becomes
$$
\sum_{i=1}^{n} i^{2} y_{1}^{2}+2 \sum_{1 \leqslant \sum_{k<j \leqslant n}} k^{2} y_{k} y_{l}=1 \text {. }
$$

Let $a_{k}=y_{k}+y_{k+1}+\cdots+y_{n}$. Then equation (1) becomes
$$
\begin{array}{l}
a_{1}^{2}+3 a_{2}^{2}+\cdots+(2 k-1) a_{k}^{2}+\cdots+(2 n-1) a_{n}^{2} \\
=1 .
\end{array}
$$

If we let $a_{n+1}=0$, then from $y_{k}=a_{k}-a_{k+1}$ we get
$$
\begin{array}{l}
\sum_{k=1}^{n} x_{k}=\sum_{k=1}^{n} k\left(a_{k}-a_{k+1}\right)=\sum_{k=1}^{n} k a_{k}-\sum_{k=1}^{n} k a_{k+1} \\
=\sum_{k=1}^{n} k a_{k}-\sum_{k=1}^{n}(k-1) a_{k}=\sum_{k=1}^{n} a_{k} .
\end{array}
$$

Therefore, by the Cauchy-Schwarz inequality and equation (2), we have
$$
\begin{array}{l}
\left(\sum_{k=1}^{n} x_{k}\right)^{2}=\left(\sum_{k=1}^{n} a_{k}\right)^{2} \\
\leqslant\left(\sum_{k=1}^{n} \frac{1}{2 k-1}\right)\left(\sum_{k=1}^{n}(2 k-1) a_{k}^{2}\right)=\sum_{k=1}^{n} \frac{1}{2 k-1} . \\
\therefore \sum_{k=1}^{n} x_{k} \leqslant \sqrt{\sum_{k=1}^{n} \frac{1}{2 k-1}} .
\end{array}
$$

The equality holds if and only if
$$
\frac{a_{1}}{1}=\frac{a_{2}}{\frac{1}{3}}=\cdots=\frac{a_{k}}{\frac{1}{2 k-1}}=\cdots=\frac{a_{n}}{\frac{1}{2 n-1}}=T \text { (a constant). }
$$

That is, $a_{k}=\frac{T}{2 k-1}(k=1,2, \cdots, n)$.
(Substituting into equation (2), we get $\left.T=\frac{1}{\sqrt{\sum_{i=1}^{n} \frac{1}{2 i-1}}}\right)$

Thus, $x_{k}=k\left(a_{k}-a_{k+1}\right)=\frac{2 k T}{4 k^{2}-1}(1 \leqslant k \leqslant n-1)$.
$$
x_{n}=n a_{n}=\frac{n T}{2 n-1} .
$$

Therefore, the maximum value of $\sum_{k=1}^{n} x_{k}$ is $\sqrt{\sum_{k=1}^{n} \frac{1}{2 k-1}}$.