9. Given an arithmetic sequence $\left\{a_{n}\right\}$ with $2 n+1$ terms, the sum of the odd-numbered terms is 2002, and the sum of the even-numbered terms is 2000. Find the middle term and the number of terms.
9. From the conditions, we know that $a_{1}+a_{3}+\cdots+a_{2 n+1}=2002$, and $a_{1}+a_{2}+\cdots+a_{2 n+1}=\frac{n+1}{2}\left(a_{1}+\right.$ $\left.a_{2 n+1}\right)=(n+1) a_{n+1}=2002$. Similarly, we can get $a_{2}+a_{4}+\cdots+a_{2 n}=\frac{n}{2}\left(a_{2}+a_{2 n}\right)=n a_{n+1}=2000$, so $a_{n+1}=2, n=1000$. Therefore, the middle term $a_{n+1}=2$, and the number of terms is 2001.