Solution:

By the binomial theorem, the coefficients of $x^2$ and $x^3$ are $\binom{2000}{2} a^2 b^{1998}$ and $\binom{2000}{3} a^3 b^{1997}.$ Setting them equal and cancelling $a^2 b^{1997}$ gives $$\binom{2000}{2} b = \binom{2000}{3} a, \qquad \text{so} \qquad b = \frac{1998}{3}\,a = 666a.$$

Since $\gcd(a, b) = 1,$ we must have $a = 1$ and $b = 666,$ so $a + b = 667.$