We see that $360 = 2^3 \cdot 3^2 \cdot 5$. So our number must have $4 \cdot 3 \cdot 2 = 24$ factors. Since our number must be odd we cannot have a $2$ in the factorization of this number. So we must have a $3$ in our factorization, and more than once. We could have $3^2 \cdot 5 \cdot 7 \cdot 11$ if we only wanted $3^2$ and not any more powers of $3$. We could have $3^3 \cdot 5^2 \cdot 7$, but that is greater than $3^2 \cdot 5 \cdot 7 \cdot 11$. Everything else is out of the question because the next factor of $24$ is $6$, which means that we will need $3^5$ which is unreasonable. So $3^2 \cdot 5 \cdot 7 \cdot 11 = \boxed{3465}$.