jmc

For $k=0,1,2,\dots ,n,$ let $P_k=(k^2,a_1+a_2+\cdots +a_k).$ Note that $P_0=(0,0)$ and $P_n=(n^2,a_1+a_2+\cdots +a_n)=(n^2,17).$



Then for each $k=1,2,\dots ,n,$ we have $$\begin{array}{rl}{P}_{k-1}{P}_k& =\sqrt{(k^2-(k-1{)}^2)+((a_1+a_2+\cdots +a_{k-1}+a_k)-(a_1+a_2+\cdots +a_{k-1}){)}^2}\\ & =\sqrt{(2k-1{)}^2+a_k^2},\end{array}$$so that $S_n$ is the minimum value of the sum $P_0{P}_1+P_1{P}_2+\cdots +P_{n-1}{P}_n.$ By the triangle inequality, $$P_0{P}_1+P_1{P}_2+\cdots +P_{n-1}{P}_n\ge P_0{P}_n=\sqrt{n^4+289}.$$Furthemore, equality occurs when all the $P_i$ are collinear, so $S_n=\sqrt{n^4+289}$ for each $n.$

It remains to find the $n$ for which $S_n$ is an integer, or equivalently, $n^4+289$ is a perfect square. Let $n^4+289=m^2$ for some positive integer $m.$ Then $m^2-n^4=289,$ which factors as $$(m-n^2)(m+n^2)=289.$$Since $n^2$ is positive and $289=17^2,$ the only possibility is $m-n^2=1$ and $m+n^2=289,$ giving $m=145$ and $n^2=144.$ Thus $n=\sqrt{144}=\boxed{12}.$