For $n \in \mathbf{N}$, let $S_{n}$ be
$$
\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}
$$

the minimum value, where $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, and $a_{1}+a_{2}+\cdots+a_{n}=17$. If there exists a unique $n$ such that $S_{n}$ is also an integer, find the value of $n$.
Solution: Based on the structure of $\sqrt{(2 k-1)^{2}+a_{k}^{2}}$, we construct the complex number $z_{k}=(2 k-1)+\sigma_{k} i, k=1,2, \cdots, n$.
Then $\sqrt{(2 k-1)^{2}+a_{k}^{2}}=\left|(2 k-i)+a_{k} i\right|$.

Thus $\sum_{k=1}^{n} \vee \sqrt{(2 k-1)^{2}+a_{k}^{2}}$
$=\sum_{k=1}^{n}\left|(2 k-1)+a_{k} i\right|$
$=\sum_{k=1}^{n}\left|z_{k}\right| \geqslant\left|\sum_{k=1}^{n} z_{k}\right|$
$=\mid[1+3+\cdots+(2 n-1)]$
$+\left(a_{1}+a_{2}+\cdots+a_{n}\right) i \mid$
$=\left|n^{2}+17 i\right|=\sqrt{n^{4}+17^{2}}$.
Given that $S_{n}$ is the minimum value of $\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}$, we have
$$
S_{n}=\sqrt{n^{4}+17^{2}} \text {. }
$$

Since $S_{n}$ is an integer, let $S_{n}=m$. Then we have
$$
\begin{array}{l}
n^{4}+17^{2}=m^{2}, \\
\left(m-n^{2}\right)\left(m+n^{2}\right)=289 . \\
\therefore m-n^{2}=1, m+n^{2}=289 .
\end{array}
$$

Solving these equations, we get $n=12$.
2 Using $|z|^{2}=z \bar{z}$, convert the modulus operation to complex number operations.