jmc

Letting $a_1=x$ and $a_2=y,$ we have $$\begin{array}{rl}{a}_3& =y-x,\\ a_4& =(y-x)-y=-x,\\ a_5& =-x-(y-x)=-y,\\ a_6& =-y-(-x)=x-y,\\ a_7& =(x-y)-(-y)=x,\\ a_8& =x-(x-y)=y.\end{array}$$Since $a_7=a_1$ and $a_8=a_2,$ the sequence repeats with period $6$; that is, $a_{k+6}=a_k$ for all positive integers $k.$

Furthermore, the sum of any six consecutive terms in the sequence equals $$x+y+(y-x)+(-x)+(-y)+(x-y)=0.$$So, since $1492$ is $4$ more than a multiple of six, the sum of the first $1492$ terms is equal to the sum of the first four terms: $$\begin{array}{rl}1985& =a_1+a_2+\cdots +a_{1492}\\ & =a_1+a_2+a_3+a_4\\ & =x+y+(y-x)+(-x)\\ & =2y-x.\end{array}$$Similarly, since $1985$ is $5$ more than a multiple of six, we have $$\begin{array}{rl}1492& =a_1+a_2+\cdots +a_{1985}\\ & =a_1+a_2+a_3+a_4+a_5\\ & =x+y+(y-x)+(-x)+(-y)\\ & =y-x.\end{array}$$Subtracting this second equation from the first equation, we get $y=1985-1492=493.$

Since $2001$ is $3$ more than a multiple of six, we have $$\begin{array}{rl}{a}_1+a_2+\cdots +a_{2001}& =a_1+a_2+a_3\\ & =x+y+(y-x)\\ & =2y=2\cdot 493=\boxed{986}.\end{array}$$(Note that solving for $x$ was not strictly necessary.)