9. In the sequence $\left\{a_{n}\right\}$, it is known that
$$
a_{1}=2, a_{n}+a_{n+1}=1\left(n \in \mathbf{N}_{+}\right) \text {. }
$$

If $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$, then the value of $S_{2003}-2 S_{2004}+S_{2000}$ is $\qquad$
9.3.

According to the problem, when $n$ is even, we have $a_{1}+a_{2}=1, a_{3}+a_{4}=1, \cdots, a_{n-1}+a_{n}=1$, a total of $\frac{n}{2}$ pairs, so $S_{n}=\frac{n}{2}$. Therefore, $S_{2004}=1002$.
When $n$ is odd, we have $a_{2}+a_{3}=1, a_{3}+a_{4}=1, \cdots, a_{n-1}+a_{n}=1$, a total of $\frac{n-1}{2}$ pairs, so
$$
S_{n}=a_{1}+\frac{n-1}{2}=2+\frac{n-1}{2}=\frac{n+3}{2} \text {. }
$$

Therefore, $S_{2003}=1003, S_{2005}=1004$.
Thus, $S_{2003}-2 S_{2004}+S_{2005}$
$$
=1003-2 \times 1002+1004=3 .
$$