2. As shown in Figure 2, the first polygon is "expanded" from an equilateral triangle, with the number of sides denoted as $a_{3}$, the second polygon is "expanded" from a square, with the number of sides denoted as $a_{4}, \cdots \cdots$ and so on. The polygon "expanded" from a regular $n(n \geqslant 3)$-sided polygon has the number of sides denoted as $a_{n}$. When $\frac{1}{a_{3}}+\frac{1}{a_{4}}+\cdots+\frac{1}{a_{n}}$ equals $\frac{2007}{6030}$, the value of $n$ is $\qquad$ .
(1)
(2)
(3)
(4)
Figure 2
2. 2009 .

From the extended definition, we know
$$
\begin{array}{l}
a_{3}=12=3 \times 4, \\
a_{4}=20=4 \times 5, \\
\cdots \cdots \\
a_{n}=n(n+1) . \\
\text { Also, } \frac{1}{a_{n}}=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} \\
\Rightarrow \frac{1}{a_{3}}+\frac{1}{a_{4}}+\cdots+\frac{1}{a_{n}} \\
=\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\cdots+\left(\frac{1}{n}-\frac{1}{n+1}\right) \\
=\frac{1}{3}-\frac{1}{n+1}=\frac{2007}{6030} \\
\Rightarrow \frac{1}{n+1}=\frac{3}{6030}=\frac{1}{2010} \Rightarrow n=2009 .
\end{array}
$$