7. Let the area of rectangle $A B C D$ be 2013, and point $E$ lies on side $C D$. Then the area of the triangle formed by the centroids of $\triangle A B E$, $\triangle B C E$, and $\triangle A D E$ is $\qquad$

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
7. $\frac{671}{3}$.

As shown in Figure 4, let the centroids of $\triangle A B E$, $\triangle B C E$, and $\triangle A D E$ be $X$, $Y$, and $Z$ respectively.

By the properties of centroids and similar triangles, it is easy to know that the distances from points $Y$ and $Z$ to side $C D$ are both $\frac{1}{3} A D$, hence $Y Z \parallel C D$.

Also, the distance from point $Y$ to side $B C$ is $\frac{1}{3} C E$, and the distance from point $Z$ to side $A D$ is $\frac{1}{3} D E$, so $Y Z = \frac{2}{3} C D$.

The distance from point $X$ to side $A B$ is $\frac{1}{3} A D$, so the distance from point $X$ to side $Y Z$ is $\frac{1}{3} A D$. Therefore,
$$
\begin{array}{l}
S_{\triangle X Y Z}=\frac{1}{2} \times \frac{2}{3} C D \times \frac{1}{3} A D \\
=\frac{1}{9} C D \cdot A D=\frac{2013}{9}=\frac{671}{3} .
\end{array}
$$