14. 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 1995, and the sum of the first 1985 terms is 1492, find the sum of the first 2001 terms.
14. $a_{n+6}=a_{n}(n \geqslant 1)$, 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 $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}=0$, so $a_{6 k+1}$ $+a_{6 k+2}+a_{6 k+3}+a_{6 k+4}+a_{6 k+5}+a_{6 k+6}=0$. By the problem, $1985=S_{1492}=2 a_{2}-a_{1}, 1492=S_{1985}=$ $a_{2}-a_{1}$, solving these gives $a_{2}=493, a_{1}=-999$. Therefore, $S_{2001}=S_{6 \times 333+3}=2 a_{2}=986$.