Consider a regular $2 n+1$-gon $P$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$, if the line segment $S E$ contains no other points that lie on the sides of $P$ except $S$. We want to color the sides of $P$ in 3 colors, such that every side is colored in exactly one color, and each color must be used at least once. Moreover, from every point in the plane external to $P$, at most 2 different colors on $P$ can be seen (ignore the vertices of $P$, we consider them colorless). Find the largest positive integer for which such a coloring is possible.
Answer: $n=1$ is clearly a solution, we can just color each side of the equilateral triangle in a different color, and the conditions are satisfied. We prove there is no larger $n$ that fulfills the requirements.

Lemma 1. Given a regular $2 n+1$-gon in the plane, and a sequence of $n+1$ consecutive sides $s_{1}, s_{2}, \ldots, s_{n+1}$ there is an external point $Q$ in the plane, such that the color of each $s_{i}$ can be seen from $Q$, for $i=1,2, \ldots, n+1$.

Proof. It is obvious that for a semi-circle $S$, there is a point $R$ in the plane far enough on the perpendicular bisector of the diameter of $S$ such that almost the entire semi-circle can be seen from $R$.

Now, it is clear that looking at the circumscribed circle around the $2 n+1$-gon, there is a semi-circle $S$ such that each $s_{i}$ either has both endpoints on it, or has an endpoint that is on the semi-circle, and is not on the semicircle's end. So, take $Q$ to be a point in the plane from which almost all of $S$ can be seen, clearly, the color of each $s_{i}$ can be seen from $Q$. $\diamond$ Take $n \geq 2$, denote the sides $a_{1}, a_{2}, \ldots, a_{2 n+1}$ in that order, and suppose we have a coloring that satisfies the condition of the problem. Let's call the 3 colors red, green and blue. We must have 2 adjacent sides of different colors, say $a_{1}$ is red and $a_{2}$ is green. Then, by Lemma 1 :

(i) We cannot have a blue side among $a_{1}, a_{2}, \ldots, a_{n+1}$.
(ii) We cannot have a blue side among $a_{2}, a_{1}, a_{2 n+1}, \ldots, a_{n+3}$.

We are required to have at least one blue side, and according to 1) and 2), that can only be $a_{n+2}$, so $a_{n+2}$ is blue. Now, applying Lemma 1 on the sequence of sides $a_{2}, a_{3}, \ldots, a_{n+2}$ we get that $a_{2}, a_{3}, \ldots, a_{n+1}$ are all green. Applying Lemma 1 on the sequence of sides $a_{1}, a_{2 n+1}, a_{2 n}, \ldots, a_{n+2}$ we get that $a_{2 n+1}, a_{2 n}, \ldots, a_{n+3}$ are all red.

Therefore $a_{n+1}, a_{n+2}$ and $a_{n+3}$ are all of different colors, and for $n \geq 2$ they can all be seen from the same point according to Lemma 1, so we have a contradiction.