Answer: 12

Small Hint:

Interpret each radical as the length of a vector $(2k-1,a_k)$ and apply the triangle inequality

Big Hint:

After finding $S_n$, factor the difference of two squares that results from requiring it to be an integer

Solution:

The two component sums are $\sum_{k=1}^n(2k-1)=n^2$ and $\sum_{k=1}^na_k=17.$ Therefore the triangle inequality for vectors gives $$\begin{aligned}&\sum_{k=1}^n\sqrt{(2k-1)^2+a_k^2}\\&\quad\geq\sqrt{(n^2)^2+17^2}\\&\quad=\sqrt{n^4+289}.\end{aligned}$$ Equality is attainable by taking $a_k$ proportional to $2k-1,$ so $S_n=\sqrt{n^4+289}.$

If this is the integer $m,$ then $$(m-n^2)(m+n^2)=289=17^2.$$ The factor pair $17,17$ gives $n=0,$ while the pair $1,289$ gives $m=145$ and $n^2=144.$ Thus the unique positive $n$ is $12.$