jmc

By inspection, we find that $$\begin{array}{rl}1\cdot 1& \equiv 1(\mathrm{mod}9)\\ 2\cdot 5& \equiv 1(\mathrm{mod}9)\\ 4\cdot 7& \equiv 1(\mathrm{mod}9)\\ 8\cdot 8& \equiv 1(\mathrm{mod}9).\end{array}$$So 1, 2, 4, 5, 7, and 8 have modular inverses (mod 9). Since no multiple of 0, 3, and 6 can be one more than a multiple of 9, we find that $\boxed{6}$ of the modulo-9 residues have inverses.