Use your inequalities!

By applying the Arithmetic-Mean Geometric-Mean (AM-GM) inequality , $\dfrac{a_{i}^2+1}{a_{i}}$ $\geq 2.$ Thus, the sum is at least $2n,$ so we know that $n=\frac{2016}{2} = \boxed{1008}$ will be sufficient for the inequality to hold.

To prove that no smaller $n$ will suffice, note that the equality $\dfrac{a_{i}^2+1}{a_{i}} = 2$ occurs where $a=1.$ Thus, if we set all $a_i = 1,$ the sum will be exactly $2n,$ so no $n < 1008$ will suffice.

By Titu’s Lemma, $\frac{a^{2} + 1} {a}$ is bigger or equal to $\frac{(a+1)^{2}}{2a}$ , which is equivalent to $\frac{a} {2} +1+\frac{1} {2a}$ . This is true for all a’s in this scenario since they are positive reals. Now, examine the approach of simplification of $\frac{a}{2} +\frac {1}{2a}$ : by AM-GM inequality, we obtain $\frac{a+ \frac{1}{a}} {2}$ is bigger or equal to $\sqrt{a \times \frac {1} {a}}$ $=1$ . Therefore, in amalgamation of the analysis, the sum of $\frac{a^{2}+1} {a}$ is bigger or equal to the sum of 2’s. In other words, $n$ terms of $a$ insinuates a minimal sum of $2n$ . Now, dividing 2016 by 2 yields 1008.