8. Given the sequence $\left.\mid a_{n}\right\}$, where $a_{n}$ is an integer, and for $n \geqslant 3, n \in \mathbf{N}$, we have $a_{n}=a_{n-1}-$ $a_{n-2}$, if the sum of the first 1985 terms is 1000, and the sum of the first 1995 terms is 4000, then the sum of the first 2002 terms is $\qquad$ _.
8. 3000

Let $a_{1}=a, a_{2}=b$, then the first 6 terms of the sequence are $a, b, b-a, -a, -b, a-b$, and $a_{7}=$ $a, a_{8}=b$. It is easy to prove that the sum of any consecutive 6 terms in this sequence is 0.
So $S_{1995}=S_{332 \times 6+3}=0+a+b+(b-a)=2 b=4000$
$$
S_{1985}=S_{330 \times 6+5}=0+a+b+(b-a)+(-a)+(-b)=b-a=1000
$$

Solving (1) and (2) gives $b=2000, a=1000$.
Then $S_{2012}=S_{333 \times 6+4}=0+a+b+(b-a)+(-a)=2 b-a=3000$.