Recall that a perfect square is the square of some integer. How many perfect squares less than 10,000 can be represented as the difference of two consecutive perfect cubes?
+179 Note that...
$1-0 = 1$
$4-1 = 3$
$9-4 = 5$
$16-9 = 7$
$25-16 = 9$
$...$
Thus, every positive odd number can be written as the difference between two perfect squares. We consider squares in the range of $1 - 99$, since $100^2 = 10,000$. An even number squared will never be odd, and vice versa an odd number squares will always be even.
Thus, taking $\frac{99+1}{2} = \boxed{50}$, which is our answer.
