15. For a positive integer $n$, let $S_{n}$ be
$$
\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}
$$

the minimum value. Where $a_{1}, a_{2}, \cdots, a_{\mathrm{n}}$ are positive integers, and their sum is 17. There is a unique value of $n$ for which $S_{n}$ is an integer, find $n$.
15. Consider each $l_{i}=\sqrt{(2 k-1)^{2}+a_{k}^{2}}$ as the hypotenuse of a right-angled triangle, with the two legs being $2 k-1$ and $a_{b}$. When these right-angled triangles are placed together to form a ladder, let $A, B$ be the starting and ending points, respectively.
The distance from $A$ to $B$ is
$$
\begin{array}{l}
\left(\sum_{k=1}^{n} a_{b}\right)^{2}+\left(\sum_{k=1}^{n}(2 h-1)\right)^{2} \\
=\sqrt{17^{2}+n^{4}}
\end{array}
$$
Let $\sum_{k=1}^{a} t_{i}=\sqrt{17^{2}+n^{4}}$. And we can choose $a_{h}$ such that $S_{n}=\sqrt{17^{2}+n^{4}}$. When $S_{n}$ is an integer, then
$$
\begin{array}{l}
17^{2}=S_{u}^{2}-n^{4}=\left(S_{n}-n^{2}\right)\left(S_{u}+n^{2}\right) . \\
\text { By }\left\{\begin{array}{l}
S_{u}+n^{2}=17^{2} \\
S_{n}-n^{2}=1,
\end{array} \text { we solve to get } S_{n}=1+5, n=12\right. \text {. }
\end{array}
$$