jmc

Any positive integer divisor of $N$ must take the form $2^a\cdot 3^b\cdot 5^c\cdot 7^d$ where $0\le a\le 4$, $0\le b\le 3$, $0\le c\le 2$, $0\le d\le 2$. In other words, there are 5 choices for $a$, 4 choices for $b$, 3 choices for $c$, and 3 choices for $d$. So there are $5\cdot 4\cdot 3\cdot 3=\boxed{180}$ natural-number factors of $N$.