11. Let the sequence of positive numbers $\left\{a_{n} \mid\right.$ have the sum of its first $n$ terms as $b_{n}$, the sequence $\left\{b_{n} \mid\right.$ have the product of its first $n$ terms as $c_{n}$, and $b_{n}+c_{n}=1$. Then the number in the sequence $\left\{\frac{1}{a_{n}}\right\}$ that is closest to 2002 is $\qquad$.
11. 1980

According to the problem, we have $b_{n}=\frac{C_{n}}{C_{n-1}}(n \geqslant 2)$ and $b_{n}+C_{n}=1$, thus $\frac{C_{n}}{C_{n-1}}+C_{n}=1$ which simplifies to $\frac{1}{C_{n}} - \frac{1}{C_{n-1}}=1$. Given $C_{1}=b, C_{1}+b_{1}=1$, we can derive $C_{1}=b_{1}=\frac{1}{2}$.
Therefore, $C_{n}=\frac{1}{n+1}, b_{n}=\frac{n}{n+1}$, and $\frac{1}{a_{n}}=n(n+1)$.
The number in the sequence $\left\{\frac{1}{a_{n}}\right\}$ closest to 2000 is $44 \times 45=1980$.