9. (12 points) For the sequence of positive numbers $\left\{a_{n}\right\}$, the sum of the first $n$ terms is $b_{n}$, and the product of the first $n$ terms of the sequence $\left\{b_{n}\right\}$ is $c_{n}$, and it is given that $b_{n}+2 c_{n}=1$ $\left(n \in \mathbf{Z}_{+}\right)$. Find the number in the sequence $\left\{\frac{1}{a_{n}}\right\}$ that is closest to 2013.
9. Since $a_{1}=b_{1}=c_{1}$, therefore, in $b_{n}+2 c_{n}=$ $\left(n \in \mathbf{Z}_{+}\right)$, let $n=1$, we get
$$
3 a_{1}=1 \Rightarrow a_{1}=\frac{1}{3} \text {. }
$$

When $n \geqslant 2$, from $b_{n}=\frac{c_{n}}{c_{n-1}}$ and $b_{n}+2 c_{n}=1$, we get $\frac{c_{n}}{c_{n-1}}+2 c_{n}=1 \Rightarrow \frac{1}{c_{n}}=\frac{1}{c_{n-1}}+2$.
Also, $\frac{1}{c_{1}}=\frac{1}{a_{1}}=3$, thus,
$$
\frac{1}{c_{n}}=3+2(n-1)=2 n+1\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

Hence, $b_{n}=\frac{c_{n}}{c_{n-1}}=\frac{\frac{1}{c_{n-1}}}{\frac{1}{c_{n}}}=\frac{2 n-1}{2 n+1}(n \geqslant 2)$.
This result also holds for $n=1$ (since $b_{1}=a_{1}=\frac{1}{3}$).
Then, $a_{n}=b_{n}-b_{n-1}=\frac{2 n-1}{2 n+1}-\frac{2 n-3}{2 n-1}$
$$
=\frac{4}{4 n^{2}-1}(n \geqslant 2) \text {. }
$$

Thus, $\frac{1}{a_{n}}=n^{2}-\frac{1}{4}(n \geqslant 2)$.
Therefore, $\frac{1}{a_{44}}=1935 \frac{3}{4}, \frac{1}{a_{45}}=2024 \frac{3}{4}$.
Upon comparison, $\frac{1}{a_{45}}$ is closest to 2013.