There are $2019$ points given in the plane. A child wants to draw $k$ (closed) discs in such a manner, that for any two distinct points there exists a disc that contains exactly one of these two points. What is the minimal $k$, such that for any initial configuration of points it is possible to draw $k$ discs with the above property?
1. **Claim that \( k \ge 1010 \):**
   - Consider 2019 collinear points on a line \(\ell\).
   - To separate any two consecutive points, each disc must intersect \(\ell\) at least once.
   - There are 2018 gaps between 2019 points, requiring at least 2018 intersections.
   - Additionally, to separate the leftmost and rightmost points, we need at least one more intersection.
   - Therefore, \( k \ge \lceil \frac{2019}{2} \rceil = 1010 \).

2. **Show that \( k = 1010 \) works:**
   - Consider a half-plane as a disc, which simplifies the problem.
   - Use inversion at an arbitrary point to ensure no three points are collinear.
   - We need to show that for any 2019 points with no three collinear, we can draw 1010 lines such that any two points are separated by some line.

3. **Lemma: For any two sets \( A \) and \( B \) of points, there is a line bisecting both \( A \) and \( B \):**
   - A line bisects a set if the numbers of points on either side differ by at most one, and no point lies on the line.
   - By a continuity argument, rotating and moving the half-plane ensures it contains half the points in \( A \) and \( B \).
   - This lemma is proven by continuity, ensuring a bisecting line exists.

4. **Partitioning the points:**
   - Start with the set \(\{P_1, P_2, \cdots, P_{2019}\}\).
   - Use a line to divide the points into sets \( A \) and \( B \) of sizes 1009 and 1010, respectively.
   - At each stage, pair up existing sets and draw a line bisecting each pair, creating four sets from two.
   - The first line increases the number of sets by 1, and each subsequent line increases the number of sets by 2, except the last one which increases by 1.
   - This process results in 1010 lines, ensuring any two points are separated by some line.

Thus, the minimal \( k \) is 1010.

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