jmc

Since $9^4=9(9^3)$ and $9^5=9^2\cdot 9^3=81(9^3)$, we must count the number of integers between 10 and 80, inclusive. That number is $80-10+1=71$, so there are $\boxed{71}$ multiples of $9^3$ greater than $9^4$ and less than $9^5$.