9. Let $f(x)$ be a function defined on $\mathbf{R}$, if $f(0)$ $=1008$, and for any $x \in \mathbf{R}$, it satisfies
$$
\begin{array}{l}
f(x+4)-f(x) \leqslant 2(x+1), \\
f(x+12)-f(x) \geqslant 6(x+5) .
\end{array}
$$

Then $\frac{f(2016)}{2016}=$ $\qquad$
9. 504 .

From the given conditions, we have
$$
\begin{array}{l}
f(x+12)-f(x) \\
=(f(x+12)-f(x+8))+ \\
\quad(f(x+8)-f(x+4))+(f(x+4)-f(x)) \\
\leqslant 2((x+8)+1)+2((x+4)+1)+2(x+1) \\
=6 x+30=6(x+5) . \\
\text { Also, } f(x+12)-f(x) \geqslant 6(x+5), \text { thus, } \\
f(x+12)-f(x)=6(x+5) .
\end{array}
$$

Also, \( f(x+12)-f(x) \geqslant 6(x+5) \), thus,

Then \( f(2016) \)
$$
\begin{array}{l}
=\sum_{k=0}^{167}(f(12 k+12)-f(12 k))+f(0) \\
=6 \sum_{k=0}^{167}(12 k+5)+1008 \\
=6 \times \frac{(2009+5) \times 168}{2}+1008 \\
=1008 \times 1008 .
\end{array}
$$

Therefore, \( \frac{f(2016)}{2016}=\frac{1008}{2}=504 \).