7. Given that on the side $C A$ of $\angle A C B$ there are 2011 points $A_{1}, A_{2}, \cdots, A_{2011}$, and on the side $C B$ there are 2011 points $B_{1}, B_{2}, \cdots, B_{2011}$, satisfying
$$
\begin{array}{l}
A_{1} A_{2}=A_{2} A_{3}=\cdots=A_{2010} A_{2011}, \\
B_{1} B_{2}=B_{2} B_{3}=\cdots=B_{2010} B_{2011} .
\end{array}
$$

If $S_{\text {quadrilateral } A_{1} A_{3} B_{3} B_{1}}=2, S_{\text {quadrilateral } A_{2} A_{4} B_{4} B_{2}}=3$, then
$$
S_{\text {quadrilateral } A_{2} A_{2011} B_{2011} B_{2009}}=
$$
. $\qquad$
7.2010 .
$$
\begin{array}{l}
\text { Let } \angle A C B=\alpha, C A_{1}=a, C B_{1}=b, \\
A_{i} A_{i+1}=s, B_{i} B_{i+1}=t(i=1,2, \cdots, 2010), \\
S_{\text {trapezoid } A_{i} A_{i+1} B_{i+2} B_{i}}=S_{i}(i=1,2, \cdots, 2009) . \\
\text { Then } S_{i}=\frac{1}{2}\{[a+(i+1) s][b+(i+1) t]- \\
\quad[a+(i-1) s][b+(i-1) t]\} \sin \alpha \\
=(a t+b s+2 i s t) \sin \alpha .
\end{array}
$$

Therefore, $\left\{S_{i}\right\}$ is an arithmetic sequence.
Given $S_{1}=2, S_{2}=3$, we get $S_{2009}=S_{2010}$.