19. For $n \in \mathbf{N}^{*}$, let $S_{n}$ be the minimum value of $\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}$, where $a_{1}, a_{2}, \cdots, a_{n} \in \mathbf{R}^{+}$, and $\sum_{k=1}^{n} a_{k}=17$. If there exists $n$ such that $S_{n}$ is also an integer, find all values of $n$.
19. From $\sqrt{(2 k-1)^{2}+a_{k}^{2}}=\left|(2 k-1)+a_{k} \mathrm{i}\right|, k=1,2, \cdots, n$, we get $\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}=$ $\sum_{k=1}^{n}\left|(2 k-1)+a_{k} i\right| \geqslant\left|\sum_{k=1}^{n}(2 k-1)+\left(\sum_{k=1}^{n} a_{k}\right) i\right|=\left|n^{2}+17 i\right|=\sqrt{n^{4}+17^{2}}$, clearly the equality can be achieved, so $s_{n}=\sqrt{n^{4}+17^{2}}>0$.
If $s_{n} \in \mathbf{Z}$, then $\left(s_{n}-n^{2}\right)\left(s_{n}+n^{2}\right)=17^{2}$, thus $s_{n}-n^{2}=1$ and $s_{n}+n^{2}=17^{2}$, solving these gives $n=12$.