jmc

Note that $$\begin{array}{rl}f(x)+f(1-x)& =\frac{3}{9^x+3}+\frac{3}{9^{1-x}+3}\\ & =\frac{3}{9^x+3}+\frac{3\cdot 9^x}{9+3\cdot 9^x}\\ & =\frac{3}{9^x+3}+\frac{9^x}{3+9^x}\\ & =\frac{3+9^x}{9^x+3}\\ & =1.\end{array}$$Thus, we can pair the 1000 terms in the sum into 500 pairs, such that the sum of the terms in each pair is 1. Therefore, the sum is equal to $\boxed{500}.$