4. When $n$ is a positive integer, it is defined that
$$
n!=n \times(n-1) \times \cdots \times 2 \times 1 \text {, }
$$

which is called the factorial of $n$ (for example, $10!=10 \times 9 \times \cdots \times 2 \times 1$ $=3628800$). Therefore, in 2010!, the total number of zeros at the end is $\qquad$
4.501.

The number of trailing zeros depends on the number of factors of 10. Since $10=2 \times 5$, and it is clear that in 2010! the number of factors of 2 is greater than the number of factors of 5, we only need to find the number of prime factors of 5 in 2010!, which gives us the number of trailing zeros.

Among the integers from 1 to 2010, there are $\left[\frac{2010}{5}\right]$ multiples of 5, $\left[\frac{2010}{5^{2}}\right]$ multiples of $5^{2}$, $\qquad$
Since $5^{5}>2010$, the number of factors of 5 in 2010! is
$$
\begin{array}{l}
{\left[\frac{2010}{5}\right]+\left[\frac{2010}{5^{2}}\right]+\left[\frac{2010}{5^{3}}\right]+\left[\frac{2010}{5^{4}}\right]} \\
=402+80+16+3=501 .
\end{array}
$$