1. Given the integer sequence $\left\{a_{n}\right\}$ satisfying $a_{n}=a_{n-1}-a_{n-2}(n \geqslant 3)$, if the sum of the first 1492 terms is 1985, and the sum of the first 1985 terms is 1492, find the sum of the first 2001 terms.
1. Observing that $\left\{a_{n}\right\}$ is a sequence with a period of 6, and using the recurrence relation, we have $a_{n+6}=a_{n}(n \geqslant 1)$, thus $a_{6 k+1}=a_{1}, a_{6 k+2}=a_{2}, a_{6 k+3}=a_{3}=a_{2}-a_{1}, a_{6 k+4}=a_{4}=-a_{1}, a_{6 k+5}=a_{5}=$ $-a_{2}, a_{6 k+6}=a_{6}=-a_{2}+a_{1}$, and $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}=0$, hence $a_{6 k+1}+a_{6 k+2}+$ $a_{6 k+3}+a_{6 k+4}+a_{6 k+5}+a_{6 k+6}=0$. Given that $1985=S_{1492}=2 a_{2}-a_{1}, 1492=S_{1985}=a_{2}$ $-a_{1}$, solving these equations yields $a_{2}=493, a_{1}=999$. Therefore, $S_{2001}=S_{6 \times 333+3}=2 a_{2}=986$.