Let there be a regular polygon of $n$ sides with center $O$. Determine the highest possible number of vertices $k$ $(k \geq 3)$, which can be coloured in green, such that $O$ is strictly outside of any triangle with $3$ vertices coloured green. Determine this $k$ for $a)  n=2019$ ;  $b)  n=2020$.
To solve this problem, we need to determine the maximum number of vertices \( k \) that can be colored green such that the center \( O \) of the regular \( n \)-sided polygon is strictly outside any triangle formed by any three green vertices. 

1. **Understanding the Geometry**:
   - A regular \( n \)-sided polygon is symmetric about its center \( O \).
   - If we color too many vertices green, there will be a triangle formed by three green vertices that contains \( O \).

2. **Key Insight**:
   - For \( O \) to be outside any triangle formed by three green vertices, the green vertices must be distributed such that no three of them form a triangle that contains \( O \).
   - This can be achieved by ensuring that the green vertices are not too close to each other.

3. **Maximum Number of Green Vertices**:
   - To avoid forming a triangle that contains \( O \), we can color at most every other vertex green.
   - This means we can color at most \(\left\lfloor \frac{n+1}{2} \right\rfloor\) vertices green.

4. **Calculations for Specific \( n \)**:
   - For \( n = 2019 \):
     \[
     k = \left\lfloor \frac{2019+1}{2} \right\rfloor = \left\lfloor \frac{2020}{2} \right\rfloor = 1010
     \]
   - For \( n = 2020 \):
     \[
     k = \left\lfloor \frac{2020+1}{2} \right\rfloor = \left\lfloor \frac{2021}{2} \right\rfloor = 1010
     \]

Conclusion:
\[
\boxed{1010}
\]