$8 \cdot 10$ Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{n}=a_{n-1}-a_{n-2}, n \geqslant 3 .
$$

If the sum of its first 1492 terms is 1985, and the sum of the first 1985 terms is 1492, find the sum of its first 2001 terms.
[Solution] For any natural number $n$, by the assumption we have
$$
\begin{array}{l}
a_{n}, a_{n+1}, a_{n+2}=a_{n+1}-a_{n}, a_{n+3}=a_{n+2}-a_{n+1}=-a_{n}, \\
a_{n+4}=a_{n+3}-a_{n+2}=-a_{n+1}, a_{n+5}=a_{n}-a_{n+1}, a_{n+6}=a_{n} .
\end{array}
$$

From this, we can also get $a_{n}+a_{n+1}+a_{n+2}+a_{n+3}+a_{n+4}+a_{n+5}=0$. Let the sum of the first $\lambda$ terms of the sequence be $S_{n}$. Since $1492=6 \times 248+4, 1985=6 \times 330+5$, we have
$$
\begin{array}{l}
1985=S_{1492}=a_{1}+a_{2}+a_{3}+a_{4}=2 a_{2}-a_{1}, \\
1492=S_{1985}=a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=a_{2}-a_{1} .
\end{array}
$$

Solving these, we get $a_{1}=-999, a_{2}=493$. Since $2001=6 \times 333+3$, we have
$$
S_{2001}=a_{1}+a_{2}+a_{3}=2 a_{2}=986 .
$$