In the plane, there are 2020 points, some of which are black and the rest are green.

For each black point, there are exactly two green points that are at a distance of 2020 from this black point.

Determine the smallest possible number of green points.

(Walther Janous)

Answer. The smallest possible number is 45 green points.
We define "green circles" as circles with a green center and a radius of 2020. Black points can only lie where exactly two green circles intersect according to the problem statement.

If we have $g$ green circles, each can intersect with every other at most twice, so there are at most $\left(\begin{array}{l}g \\ 2\end{array}\right) \cdot 2=g^{2}-g$ black points (or even fewer if more than two green circles intersect at some of these points). Adding the $g$ green points, there are thus at most $g^{2}-g+g=g^{2}$ points in total.

Since $44^{2}<2020\left(<45^{2}\right)$, we need at least 45 green points.

We now try to place 45 green circles such that each intersects with every other twice and no point is where three or more circles intersect.

One possible construction is to choose a segment $A B$ somewhere in the plane with a length less than 4040 and place the centers of the green circles arbitrarily on it.

It is clear that any two circles intersect twice because the distance between their centers is less than twice the radius.

Furthermore, it is easy to verify that no three circles intersect at a single point, as the perpendicular distance of an intersection point to $A B$ is smaller the farther apart the centers are. If there were a point $S$ through which three circles $k_{1}, k_{2}$, and $k_{3}$ pass, then for their centers, $M_{1} M_{2}=M_{2} M_{3}=M_{1} M_{3}$ would have to hold, which is not possible since the centers all lie on a straight line.

Thus, we get a total of $\frac{45 \cdot(45-1)}{2} \cdot 2=1980$ pairwise intersection points of the 45 circles. If we color all but five of them black, we obtain the desired configuration with $45+(1980-5)=$ 2020 points.

A construction with 45 green points is therefore indeed possible, making this the smallest possible number.

(Walther Janous, Birgit Vera Schmidt)