Interesting. The following proof is a little sketchy but shouldn't be too difficult to make rigorous. Claim: $\Big\lfloor{\frac{n+1}{4}} \Big\rfloor$ colors are the maximum for $n\ge 3$. For $n\le 3$, $1$ color is trivially the maximum. Lower Bound: Mark 3 points, $A$, $B$, $C$ on a line in that order such that $AB=1$ and $BC=2$. Mark 2 positions $X,Y$ such that $X, B, Y$ collinear, $XY\perp AC$, and $XB = BY = 1$. Draw an arc of the circle centered at $B$ of radius very slightly greater than $1$ of very small ($\ll 1$) measure, centered around $X$. Mark positions $x_1, x_2, ..., x_i$ on this arc. $(i= \Big\lfloor{\frac{n+1}{4}} \Big\rfloor -1)$. Reflect these positions across $B$ to $y_1, y_2, ..., y_i$ respectively. Put pairs of points on each of these positions, such that they are an extremely small distance away (relative to the separation of the positions) from the position they are at. Note that the closest and furthest points to $A$ are $B$ and $C$ respectively, those of $B$ are $A$ and $C$ respectively, and those of $C$ are $B$ and $A$ respectively. Hence, we may assign $A$, $B$, and $C$ a color. For the points near the marked positions, the closest point will be one near the same position that it is at, while the furthest point will be one that's near the position diametrically opposite, as their distance to $C$ is very nearly $\sqrt{3}<2$. Hence for each pair of positions, we can assign them a unique color unique to the one of $A$, making $\Big\lfloor{\frac{n+1}{4}} \Big\rfloor$ colors in total. Upper Bound: Clearly for any point, the furthest and nearest points must be of the same color as it, so each color has at least $3$ points of that color, necessarily. It hence suffices to show (for $n\ge 3$) that there cannot be $2$ colors with $3$ or points that are of that color. Suppose we have points $A$, $B$, $C$, $A'$, $B'$, $C'$ such that the first and last three are all of the same color, and $AC>AB>BC, A'C'>A'B'>B'C'$, without loss of generality of course. Note that $AB'>AB$, as $B$ is the closest point to $A$, and $A'B'>AB'$ as $A'$ is the furthest point from $B'$. Hence $A'B'>AB$. However, $A'B<AB$, and $A'B'<A'B$ similarly, so $A'B'<AB$, a contradiction.