Suppose $k$ is less than $44$. Choose a point $P$ on the boundary of the convex hull of this set of $2019$ points.
By the pigeonhole principle, since $\left\lfloor \frac{2018}{43} \right\rfloor > 45$, there is a circle centered at $P$ on which at least $45$ of the other points lie. Moreover, these $45$ points all lie on the same semicircle since $P$ is on the boundary of the convex hull. Let these points be $A_1, A_2, \dots, A_{45}$ in that order. Then the distances
$$A_1A_2 < A_1A_3 < \dots < A_1A_{45}$$
are distinct, which contradicts the assumption that there are less than $44$ distances. Thus, $k \ge 44$.