4. Given the function $f(x)$ satisfies for all real numbers $x, y$, 
$$
\begin{array}{l}
f(x)+f(2 x+y)=f(3 x-y)+x-2010 \text {. } \\
\text { Then } f(2010)=
\end{array}
$$
4.0.

Let $x=2010, y=1005$, then
$$
\begin{array}{l}
f(2010)+f(5025) \\
=f(5025)+2010-2010 .
\end{array}
$$

Therefore, $f(2010)=0$.