9. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and
$$
f(x)=f(1-x) \text {. }
$$

Then $f(2010)=$ $\qquad$ .
$$
\begin{array}{l}
f(0)=0, \\
f(x+1)=f(-x)=-f(x) .
\end{array}
$$

From the conditions, we have $f(x+2)=f(x)$, which means $f(x)$ is a periodic function with a period of 2.
Therefore, $f(2010)=f(0)=0$.