First consider the case $a, b>0$. We have $720=2^{4} \cdot 3^{2} \cdot 5$, so the number of divisors of 720 is $5 * 3 * 2=30$. We consider the number of ways to select an ordered pair $(a, b)$ such that $a, b, a b$ all divide 720. Using the balls and urns method on each of the prime factors, we find the number of ways to distribute the factors of 2 across $a$ and $b$ is $\binom{6}{2}$, the factors of 3 is $\binom{4}{2}$, the factors of 5 is $\binom{3}{2}$. So the total number of ways to select $(a, b)$ with $a, b, a b$ all dividing 720 is $15 * 6 * 3=270$. The number of ways to select any $(a, b)$ with $a$ and $b$ dividing 720 is $30 * 30=900$, so there are $900-270=630$ ways to select $a$ and $b$ such that $a, b$ divide 720 but $a b$ doesn't. Now, each $a, b>0$ corresponds to four solutions $( \pm a, \pm b$ ) giving the final answer of 2520. (Note that $a b \neq 0$.)