jmc

Let $n$ be the other integer, so $$\frac{\mathrm{lcm}[45,n]}{\gcd (45,n)}=33.$$We know that $\gcd (m,n)\cdot \mathrm{lcm}[m,n]=mn$ for all positive integers $m$ and $n$, so $$\gcd (45,n)\cdot \mathrm{lcm}[45,n]=45n.$$Dividing this equation by the previous equation, we get $$[\gcd (45,n){]}^2=\frac{45n}{33}=\frac{15n}{11},$$so $11[\gcd (45,n){]}^2=15n$.

Since 11 divides the left-hand side, 11 also divides the right-hand side, which means $n$ is divisible by 11. Also, 15 divides the right-hand side, so 15 divides the left-hand side, which means $\gcd (45,n)$ is divisible by 15. Since $45=3\cdot 15$, $n$ is divisible by 15. Hence, $n$ must be divisible by $11\cdot 15=165$.

Note that $\gcd (45,165)=15$ and $\mathrm{lcm}[45,165]=495$, and $495/15=33$, so $n=165$ is achievable and the smallest possible value of $n$ is $\boxed{165}$.