Since $n$ is divisible by $20$, we can write $n = 2^{2+a} \cdot 5^{1+b} \cdot k$, where $k$ is a positive integer divisible neither by $2$ nor $5$, while $a$ and $b$ are non-negative integers. If $n^2 = 2^{2(2+a)} \cdot 5^{2(1+b)} \cdot k^2$ is a perfect cube, then $3$ divides $2(2+a)$ and $3$ divides $2(1+b)$. If $n^3 = 2^{3(2+a)} \cdot 5^{3(1+b)} \cdot k^3$ is a perfect square, then $2$ divides $3(2+a)$ and $2$ divides $3(1+b)$. From this it follows that $6$ divides $2+a$ and $6$ divides $1+b$, so $a$ is at least $4$ and $b$ is at least $5$. The smallest possible positive integer $k$ is $k=1$. When this is the case, we get $n = 2^6 \cdot 5^6 = 1000000$. We see that $n^2$ is indeed a perfect cube and $n^3$ is a perfect square, and we have shown this $n$ to be the smallest possible.