Solution:
Answer: $\frac{1}{2}$
Let $ab = 2016$. Then
$$\frac{a^{2}}{a^{2}+2016} + \frac{b^{2}}{b^{2}+2016} = \frac{a^{2}}{a^{2}+2016} + \frac{\left(\frac{2016}{a}\right)^{2}}{\left(\frac{2016}{a}\right)^{2}+2016} = \frac{a^{2}}{a^{2}+2016} + \frac{2016}{a^{2}+2016} = 1$$
Thus, every divisor $d$ pairs up with $\frac{2016}{d}$ to get $1$, so our desired expected value is $\frac{1}{2}$.