19. Consider the partition of plane $\pi$ into regular hexagons, each having inradius 2. Fix one of these hexagons, denoted by $\gamma$. For any other hexagon $x$ in the partition, there exists a unique translation $\tau_{x}$ taking it onto $\gamma$. Define the mapping $\varphi: \pi \rightarrow \gamma$ as follows: If $A$ belongs to the interior of a hexagon $x$, then $\varphi(A)=\tau_{x}(A)$ (if $A$ is on the border of some hexagon, it does not actually matter where its image is). The total area of the images of the union of the given circles equals $S$, while the area of the hexagon $\gamma$ is $8 \sqrt{3}$. Thus there exists a point $B$ of $\gamma$ that is covered at least $\frac{S}{8 \sqrt{3}}$ times, i.e., such that $\varphi^{-1}(B)$ consists of at least $\frac{S}{8 \sqrt{3}}$ distinct points of the plane that belong to some of the circles. For any of these points, take a circle that contains it. All these circles are disjoint, with total area not less than $\frac{\pi}{8 \sqrt{3}} S \geq 2 S / 9$. Remark. The statement becomes false if the constant $2 / 9$ is replaced by any number greater than $1 / 4$. In that case a counterexample is, for example, a set of unit circles inside a circle of radius 2 covering a sufficiently large part of its area.