16.14

$\left \lfloor \frac{n(n+1)(n+2)(n+3)(n+4)}{n^9 +1}\right \rfloor=\begin{cases}60 & n=1\\1 & n=2\\0&n<1,n>2\end{cases}$

Hence, the sum is simply $60+1=61$ .

$\displaystyle \large \sum_{n=0}^{2016}\left \lfloor \frac{n(n+1)(n+2)(n+3)(n+4)}{n^9 +1}\right \rfloor \\ = \displaystyle \large \sum_{n=0}^2\left \lfloor \frac{n(n+1)(n+2)(n+3)(n+4)}{n^9 +1}+ \sum_{n=3}^{2016} \frac{n(n+1)(n+2)(n+3)(n+4)}{n^9 +1}\right \rfloor \\ = \displaystyle \left \lfloor \dfrac 0 1+ \dfrac {5!} {1^9+1}+ \dfrac{\frac{6!} 2}{ 2^9+ 1} + \sum_{n=3}^{2016} \frac{n(n+1)(n+2)(n+3)(n+4)}{n^9 +1}\right \rfloor \\ =\left \lfloor 0+60+1.4035~+~ summation~much~ less~than~.5 \right \rfloor.\\ = \Large \color{#D61F06}{61}.$