Let S(1) = a and S(2) = b. Then S(3) = b-a, S(4) = -a, S(5) = -b, S(6) = a-b, S(7) = a, S(8) = b, etc. The sequence is cyclic mod 6, also the sum of any 6*n consecutive terms is zero.

Sum{S(1) to S(1492)} = S(1) + S(2) + S(3) + S(4) + Sum{S(5) to S(1492)} = a + b + b-a + -a + 0 = 2b-a = 1985

Sum{S(1) to S(1985)} = S(1) + S(2) + S(3) + S(4) + S(5) + Sum{S(6) to S(1985)} = a + b + b-a + -a + -b + 0 = b-a = 1492

Then b=493 and a = -999

Sum{S(1) to S(2001)} = S(1) + S(2) + S(3) + Sum{S(4) to S(2001)} = a + b + b-a + 0 = 2b = 986