Let $N$ be the number of complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$.

To solve the problem, we need to find the number of complex numbers \( z \) such that \( |z| = 1 \) and \( z^{720} - z^{120} \) is a real number. Let's break down the solution step by step.

1. **Expressing the condition in terms of complex conjugates:**
   Since \( |z| = 1 \), we have \( \overline{z} = \frac{1}{z} \). Therefore, the condition \( z^{720} - z^{120} \) being real can be written as:
   \[
   z^{720} - z^{120} = \overline{z^{720} - z^{120}} = \frac{1}{z^{720}} - \frac{1}{z^{120}} = \frac{z^{120} - z^{720}}{z^{840}}
   \]

2. **Simplifying the equation:**
   This implies:
   \[
   z^{840}(z^{720} - z^{120}) = z^{120} - z^{720}
   \]
   Rearranging, we get:
   \[
   z^{840} (z^{720} - z^{120}) + z^{720} - z^{120} = 0
   \]
   Factoring out \( z^{720} - z^{120} \), we have:
   \[
   (z^{840} + 1)(z^{720} - z^{120}) = 0
   \]

3. **Solving the factored equation:**
   This gives us two cases:
   - \( z^{720} = z^{120} \)
   - \( z^{840} = -1 \)

4. **Case 1: \( z^{720} = z^{120} \):**
   This implies:
   \[
   z^{120}(z^{600} - 1) = 0
   \]
   Since \( z \neq 0 \), we have \( z^{600} = 1 \). The solutions to \( z^{600} = 1 \) are the 600th roots of unity, which are:
   \[
   z = e^{2\pi i k / 600} \quad \text{for} \quad k = 0, 1, 2, \ldots, 599
   \]
   Thus, there are 600 solutions in this case.

5. **Case 2: \( z^{840} = -1 \):**
   The solutions to \( z^{840} = -1 \) are:
   \[
   z = e^{(2k+1)\pi i / 840} \quad \text{for} \quad k = 0, 1, 2, \ldots, 839
   \]
   Thus, there are 840 solutions in this case.

6. **Checking for overlaps:**
   We need to check if there are any overlaps between the solutions of \( z^{600} = 1 \) and \( z^{840} = -1 \). The solutions to \( z^{600} = 1 \) are:
   \[
   z = e^{2\pi i k / 600} = e^{i\pi k / 300}
   \]
   The solutions to \( z^{840} = -1 \) are:
   \[
   z = e^{(2\ell + 1)\pi i / 840}
   \]
   For these to be equal, we need:
   \[
   \frac{k}{300} = \frac{2\ell + 1}{840}
   \]
   Simplifying, we get:
   \[
   840k = 300(2\ell + 1) \implies 14k = 5(2\ell + 1)
   \]
   This implies:
   \[
   14k = 10\ell + 5 \implies 14k - 10\ell = 5
   \]
   This is a Diophantine equation. For \( k \) and \( \ell \) to be integers, \( 14k - 10\ell = 5 \) must have integer solutions. However, since 14 and 10 are both even, the left-hand side is always even, while the right-hand side is odd, which is a contradiction. Therefore, there are no integer solutions, and hence no overlaps.

7. **Counting the total number of solutions:**
   Since there are no overlaps, the total number of solutions is:
   \[
   N = 600 + 840 = 1440
   \]

8. **Finding the remainder when \( N \) is divided by 1000:**
   \[
   1440 \mod 1000 = 440
   \]

The final answer is \(\boxed{440}\).