Function of this year!

Substituting $n$ with $n-1$ in condition (b), we get

$\displaystyle \sum_{i=1}^{n-1} f(i) = (n-1)^2 f(n-1)$

Subtracting this from (b), we get

$\begin{aligned} f(n) &= n^2 f(n) - (n-1)^2 f(n-1)\\ (n^2-1) f(n) &= (n-1)^2 f(n-1)\\ f(n) &= \dfrac {n-1}{n+1} f(n-1) \end{aligned}$

Applying this iteratively, we get

$\begin{aligned} f(n) &= \dfrac {n-1}{n+1} \cdot \dfrac {n-2}{n} \cdot \ldots \cdot \dfrac {2}{4} \cdot \dfrac {1}{3} \cdot f(1)\\ &= \dfrac {2}{n(n+1)} f(1) \end{aligned}$

Thus, $f(2016) = \frac {2}{2016 \times 2017} \times 2016 = \frac {2}{2017}$ . Therefore, the answer is $2+2017=\boxed{2019}$ .