HMMT February

Solution:
Suppose $n=3^{a} m$, where $3 \nmid m$. Then
$$\operatorname{lcm}(n, 9)=3^{\max (a, 2)} m$$
In order for this to be a square, we require $m$ to be a square, and $a$ to either be even or $1$. This means $n$ is either a square (if $a$ is even) or of the form $3 k^{2}$ where $3 \nmid k$ (if $a=1$).

There are $31$ numbers of the first type, namely
$$1^{2}, 2^{2}, 3^{2}, 4^{2}, \ldots, 30^{2}, 31^{2}$$
There are $12$ numbers of the second type, namely
$$3 \cdot 1^{2}, 3 \cdot 2^{2}, 3 \cdot 4^{2}, 3 \cdot 5^{2}, \ldots, 3 \cdot 16^{2}, 3 \cdot 17^{2}$$
Overall, there are $31+12=43$ such $n$.