We will show that $\sum a_{n,k}\geq 2016\sqrt{n}$ by induction, which will solve the problem. The base case of $n=0$ is clearly true. For the inductive step, note that $\sum a_{n+1,k}=\sum a_{n,k}+\frac{1}{2a_{n,k}}$. The function $f(x)=x+\frac{1}{2x}$ is convex on $\mathbb{R}^+$ since $f'(x)=1-\frac{1}{2x^2}$ and $f''(x)=\frac{1}{x^3}$. Hence by Jensen, $\sum a_{n+1,k}\geq n(\sqrt{n}+\frac{1}{2\sqrt n})=2016(\sqrt n+\frac1{2\sqrt n})$. It remains to show that $\sqrt n+\frac1{2\sqrt n}\geq\sqrt{n+1}$, which after squaring becomes $n+\frac{1}{4n}+1\geq n+1$, which is true.

Just simplify the induction. Let $b_n=\sum_{k=1}^{2016} a_{n,k},$ and $a_{2017}=a_1.$ Cauchy-Swartz gives: $$\left(\sum_{k=1}^{2016} \frac{1}{2a_{n,k}} \right) \left( 2b_n \right)= \left(\sum_{k=1}^{2016} \frac{1}{2a_{n,k}} \right) \left(\sum_{k=1}^{2016} 2a_{n,k} \right) \ge 2016^2.$$Hence, $$b_{n+1}=\sum_{k=1}^{2016} a_{n+1,k} =\sum_{k=1}^{2016} \left( a_{n,k}+\frac{1}{2a_{n,k+1}}\right)=\sum_{k=1}^{2016} a_{n,k}+\sum_{k=1}^{2016} \frac{1}{2a_{n,k}} \ge b_n+\frac{2016^2}{2b_n}.$$Thus, $$b_{n+1}^2 \ge \left(b_n+ \frac{2016^2}{2b_n} \right)^2 \ge b_n^2+2016^2.$$Hence, $$b_{2016}^2 \ge b_0^2+2016^3 \implies \max_{1\leq k \leq 2016} a_{2016,k} \ge \frac{1}{2016} \sum_{k=1}^{2016} a_{2016,k}=\frac{1}{2016} b_{2016} \ge \sqrt{2016} >44.$$