Solution:

We have $N := 1! \cdot 2! \cdot 3! \cdots 10! = 2^{38} 3^{17} 5^{7} 7^{4}$. Thus, a positive divisor of $N$ that is a perfect cube must be of the form $2^{3a} 3^{3b} 5^{3c} 7^{3d}$ for some nonnegative integers $a, b, c, d$.

We see that $3a \leq 38$, $3b \leq 17$, $3c \leq 7$ and $3d \leq 4$. Thus, there are $\left\lceil \frac{38}{3} \right\rceil = 13$ choices for $a$, $\left\lceil \frac{17}{3} \right\rceil = 6$ choices for $b$, $\left\lceil \frac{7}{3} \right\rceil = 3$ choices for $c$, and $\left\lceil \frac{4}{3} \right\rceil = 2$ choices for $d$.

Hence, there are $13 \cdot 6 \cdot 3 \times 2 = 468$ positive perfect cube divisors of $N$.