jmc

A perfect square that is a multiple of $20=2^2\cdot 5^1$ must be a multiple of $2^2\cdot 5^2=100$. A perfect cube that is a multiple of 20 must be a multiple of $2^3\cdot 5^3=1000$. Our goal is thus to count the multiples of 20 from 100 to 1000 inclusive: $$100\le 20n\le 1000.$$Dividing this entire inequality by 20 we get $5\le n\le 50$, so there are $50-5+1=\boxed{46}$ integers in Cameron's list.