16. 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 Problem)
16. Notice that $\left(\sum_{i=1}^{n} x_{i}\right)^{2}=\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leq k<j \leq n} x_{k} x_{j} \geqslant \sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant k<j \leq n} \sqrt{\frac{k}{j}} x_{k} x_{j}=1$.

When $x_{i}=1, x_{k}=0(k \neq i)$, equality holds.
Let $x_{j}=\sqrt{j} y_{j}$, then $1=\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant k<j \leqslant n} \sqrt{\frac{k}{j}} x_{k} x_{j}=\sum_{i=1}^{n} i y_{i}^{2}+2 \sum_{1 \leqslant k<j \leqslant n} k y_{k} y_{i}$, i.e.,
$\left(\sum_{i=1}^{n} y_{i}\right)^{2}+\left(\sum_{i=2}^{n} y_{i}\right)^{2}+\cdots+\left(\sum_{i=n-1}^{n} y_{i}\right)^{2}+y_{n}^{2}=1$.
Let $u_{1}=\sum_{i=1}^{n} y_{i}, u_{2}=\sum_{i=2}^{n} y_{i}, \cdots, u_{n}=y_{n}$, then
$\sum_{i=1}^{n} u_{i}^{2}=1, x_{k}=\sqrt{k} y_{k}=\sqrt{k}\left(u_{k}-u_{k+1}\right)$, where $u_{n+1}=0$.
By Abel's transformation, $\left(\sum_{k=1}^{n} x_{k}\right)^{2}=\left[\sum_{k=1}^{n} \sqrt{k}\left(u_{k}-u_{k+1}\right)\right]^{2}=\left[\sum_{k=1}^{n} u_{k}(\sqrt{k}-\sqrt{k-1})\right]^{2} \leqslant \sum_{k=1}^{n} u_{k}^{2} \cdot \sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}$,
i.e., $\sum_{i=1}^{n} x_{i} \leqslant \sqrt{\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}}$, equality holds if $u_{k}=\lambda(\sqrt{k}-\sqrt{k-1})$,
Substituting into $\sum_{i=1}^{n} u_{i}^{2}=1$, we get $\lambda=\frac{1}{\sqrt{\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}}}$,
Thus, $u_{k}=\frac{\sqrt{k}-\sqrt{k-1}}{\sqrt{\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}}}$.