Find the maximum integral value of $k$ such that $0 \le k \le 2019$ and $|e^{2\pi i  \frac{k}{2019}} - 1|$ is maximal.
1. We start with the expression \( e^{2\pi i \frac{k}{2019}} \). This represents a point on the unit circle in the complex plane, where the angle is \( \frac{2\pi k}{2019} \) radians.

2. We need to find the value of \( k \) such that \( |e^{2\pi i \frac{k}{2019}} - 1| \) is maximal. This distance is maximized when \( e^{2\pi i \frac{k}{2019}} \) is as far from 1 as possible on the unit circle.

3. The point on the unit circle that is farthest from 1 is -1, which corresponds to an angle of \( \pi \) radians. Therefore, we need \( \frac{2\pi k}{2019} \) to be as close to \( \pi \) as possible.

4. Setting \( \frac{2\pi k}{2019} = \pi \), we solve for \( k \):
   \[
   \frac{2\pi k}{2019} = \pi \implies 2k = 2019 \implies k = \frac{2019}{2} = 1009.5
   \]

5. Since \( k \) must be an integer, we consider the two closest integers to 1009.5, which are 1009 and 1010.

6. We need to check which of these values maximizes \( |e^{2\pi i \frac{k}{2019}} - 1| \). Both \( k = 1009 \) and \( k = 1010 \) are equally close to \( \frac{2019}{2} \), but we are asked for the maximum possible value of \( k \).

7. Therefore, the maximum integral value of \( k \) is \( 1010 \).

\(\blacksquare\)

The final answer is \( \boxed{1010} \)