jmc

Any positive integer divisor of $N$ must take the form $2^a\cdot 3^b\cdot 5^c$ where $0\le a\le 3$, $0\le b\le 2$ and $0\le c\le 1$. In other words, there are 4 choices for $a$, 3 choices for $b$ and 2 choices for $c$. So there are $4\cdot 3\cdot 2=\boxed{24}$ natural-number factors of $N$.