Marking the point closest to each point We have $6000$ points in the plane. All distances between every pair of them are distinct.
For each point, we mark red the point nearest to it. What is the smallest number of points that can be marked red?
I divide the $6000$ points into $1000$ groups with $6$ points in each group. In each group I have one point being the center, and the other five points forming a regular pentagon around it. Then only the center points are marked red, for a total of $1000$ points. Note that using a regular hexagon is not possible, since the points will have equal distance, and using a heptagon or more will yield points other than the center being marked red.

This is a known problem, and I use its standard dramatic interpretation to solve it.  Consider $n$ marksmen standing in a field (so that all their pairwise distances are different). Each marksman simultaneously shoots and kills the closest marksman. What is the smallest number $k(n)$ of marksmen that can be killed?
Lemma 1. Each killed marksman has no more than 5 bullets in his body.
Proof. Suppose that the marksman located at the point $O$ has $m\ge 6$ bullets in his body. Suppose that these bullets belongs to marksmen located at the points $A_1,\dots, A_m$. Draw the rays $OA_1,\dots,OA_m$. Then there exist points $A_i$ and $A_j$, $i\not =j$ such that the angle $A_iOA_j\le \pi/3$. Therefore the angle $A_iOA_j$ is not the largest angle of the triangle $A_iOA_j$. Therefore the side $A_iA_j$ is not the largest side of the triangle $A_iOA_j$. But this is impossible, since, by the problem's conditions, $|A_iA_j|>|A_iO|$ and $|A_jA_i|>|A_jO|$.$\square$
Lemma 2. $k(n)\ge n/5$ for each $n$.
Proof. Count the number $b$ of the shot bullets by two ways. At the first, clearly, that $b=n$. From the other side, by Lemma 1, $5k(n)\ge b$. This implies the lemma.$\square$
