3. We define coloring of the plane as follows:

- choose a natural number $m$,
- let $K_{1}, K_{2}, \ldots, K_{m}$ be different circles with non-zero radii such that $K_{i} \subset K_{j}$ or $K_{j} \subset K_{i}$ for $i \neq j$,
- points in the plane that are outside any of the chosen circles are colored differently from the points that are inside the circle.

In the plane, there are 2019 points such that no three of them are collinear. Determine the maximum number of colors with which the given points can be colored?
Solution. From the condition of the problem, it follows that the maximum number of colors is less than or equal to 2019.

We will prove that the number 2019 colors can be achieved. For this purpose, it is sufficient to prove that there exist $K_{1}, K_{2}, \ldots, K_{2019}$ circles that define different colorings of the points. We draw all the segments that have endpoints at the given 2019 points, a total of $\frac{2019 \cdot 2018}{2}$ segments. We draw the perpendicular bisectors of these segments. We choose a point $O$ that does not lie on any of the drawn perpendicular bisectors. Clearly, the distances from this point to the given 2019 points are all different (the chosen point does not lie on any of the perpendicular bisectors of the segments) and we arrange these distances in increasing order $0<s_{1}<s_{2}<\ldots<s_{2019}$. Further, we find numbers $r_{i}, i=1,2, \ldots, 2019$ such that

$$
s_{1}<r_{1}<s_{2}<r_{2}<\ldots<s_{2019}<r_{2019}
$$

Finally, we construct 2019 concentric circles with center at point $O$ and radii $r_{i}, 1 \leq i \leq 2019$. Clearly, these circles define a coloring in which each of the selected 2019 points is colored differently.