16.9

$\displaystyle\sum_{n=0}^{2016}\left\lfloor \dfrac{n^2}{n+1}\right\rfloor \implies \displaystyle\sum_{n=0}^{2016}\left\lfloor \dfrac{n^2+n-n}{n+1}\right\rfloor \implies \displaystyle\sum_{n=0}^{2016}\left\lfloor \dfrac{n(n+1)-n}{n+1}\right\rfloor \implies \displaystyle\sum_{n=0}^{2016}\left\lfloor n- \dfrac{n}{n+1}\right\rfloor \\ = \displaystyle\sum_{n=0}^{2016}n-\displaystyle\sum_{n=0}^{2016}\left\lfloor -\dfrac{n}{n+1}\right\rfloor \\ = \dfrac{2016\cdot2017}{2}-2016 \quad \quad \left( n<n+1 \implies \dfrac{n}{n+1}<1 \right) \\ = \boxed{2031120}$