The function for 2016

Note that $(n+1)^2f(n+1)-n^2f(n)=f(n+1)$

$(n^2+2n+1-1)f(n+1)=n^2f(n)$

$f(n+1)=\frac{n^2}{n(n+2)}f(n)$

$f(n+1)=\frac{n}{n+2}f(n)$

$f(n)=\frac{n-1}{n+1}f(n-1)$

$\therefore f(n)=\frac{n-1}{n+1} \times \frac{n-2}{n} \times \frac{n-3}{n-1} \times\cdots \times \frac{2}{4} \times \frac{1}{3} \times f(1)$

Cancelling like terms and substituting $f(1)=2016$ , we get

$f(n)=\frac{2 \times 2016}{n(n+1)}$

$\therefore f(2016)=\frac{2 \times 2016}{2016(2016+1)}=\frac{2}{2017}$

Required answer: $2+2017=\boxed{2019}$