Here's a somewhat convoluted solution. We claim that the answer is $\boxed{1010}.$ We will first show that $k \ge 1010.$ Indeed, consider simply taking $2019$ collinear points, say they all lie on $\ell$. The discs must intersect $\ell$ a total of at least $2018$ times, as otherwise there are two consecutive points which are not separated by any disc. However, observe that there must be at least one more intersection point, as otherwise the leftmost and rightmost points on $\ell$ are not separated by any disc. Hence, we have $k \ge 1010.$ Now, let's show that $k = 1010$ actually works. First, we will consider a half-plane a disc. This doesn't change the problem because we could just take an arbitrarily large disk instead. Now, invert at an arbitrary point in the plane so that no three of the $2019$ points are collinear. This is doable, as we just need the center of inversion to not be concyclic with any three of the $2019$ points (which is easy to ascertain). After this, it suffices to show that for any $2019$ points with no three collinear, we can draw $1010$ lines (none of which contains any of the points) so that any two of the $2019$ points are separated by some line. Say that a line bisects a set if the numbers of points of the set on either side of the line differ by at most one, and no point of the set lies on the line. The following lemma will prove useful. Lemma. For any two sets $A, B$ of points, there is a line bisecting both $A$ and $B.$ Proof. Equivalently, we will show that there is an open half-plane containing half of the points in $A$ and half of the points in $B$, where we say that a half-plane $P$ contains half the points in $A$ if $\left| 2 \cdot |A \cap P| - |A| \right | \le 1.$ This is pretty easy to show by a continuity argument. Keep rotating and moving the half-plane such that it always contains half the points in $A$. Once it returns to the complement of the original half-plane, a continuity argument finishes. $\blacksquare$ We can now finish the problem. We will keep track of a partition $S_1 \cup S_2 \cup \cdots \cup S_k$ of the $2019$ points, call them $P_1, P_2, \cdots, P_{2019}.$ To begin, we consider the single set $\{P_1, P_2, \cdots, P_{2019}\}.$ Next, we will use a line to divide the points into sets $A$ and $B$ of size $1009, 1010$ respectively. At each stage thereafter, we will pair up the existing sets. Then, we will draw a line which bisects each pair of them, and replace the two sets with four sets according to how the line split them up. Note that the first line increases the number of sets by $1$, and each line thereafter increases the number of sets by $2$, except the last one which increases the number of sets by $1.$ This means that since the number of sets increases by $2018$, there are $1010$ lines drawn in this operation. So $1010$ works, and the answer is $\boxed{1010}.$ $\square$

The answer is $k=1010$ To show that $k\ge 1010$ just take $2019$ points on a line . Now we prove that $1010$ circles are enough to separate each pair of points. Invert from a random point such that no three inverted points are collinear. It is pretty straightforward to figure out that the following is sufficient to solve the initial problem: We are able to separate each pair of points among $2018$ points (no three collinear) using $1009$ lines . (We use one more line for the $2019th$ point which we choose to be on the convex hull) The algorithm which draws $n$ lines that separate each pair of points among $2n$ points (no three points collinear): $1)$Draw a line $l$ that bisects the set of $2n$ points , $n$ blue points and $n$ red points from each side. $2)$Repeat the following : Given $m$ blue and red points take their convex hull and choose a segment between a blue and red point that intersects $l$. Move the segment slightly so that you get a line that separates the two points from the rest.Remove the two points so that $m-1$ blue and red points remain. Do this untill only $1$ red and blue points remain and you have $n$ lines that separate the set of $2n$ points.