Moving points in a plane divided by triangles I have an infinite plane divided into triangles (as shown in the picture). I choose an equilateral triangle (as example: marked red) made of $n^2$ little triangles and place $n$ points (marked pink) in them (at most one point in each). I can move them as shown in blue. For what $n$ there exist such arrangement of points so it is possible to move then points so that all of them are in one small triangle? I think it is possible for every $n$ different than $2$, but can't prove it.

 A: It is indeed true for all $n\geq 2$. $n=3$ and $n=4$ are easy to verify by example so assume $n\geq 5$.
Let the triangles that border the right side be labelled in order $r_1,\dots, r_n$ such that $r_1$ is at the top, and similarly label the triangles that border the left side be labelled $l_1,\dots,l_n$. Then $r_1=l_1$. add a dot to each $r_{2k-1}$ and $l_{2k-1}$ for $1\leq k\leq \lceil n/2\rceil$ This gives us $2\lceil n/2\rceil-1$ dots. If $n$ is odd, this is sufficient. If $n$ is even, then we may add a dot in the center triangle of the fourth row from top. (the triangle is pointing down.)
It isn't hard to see that all of these dots can be moved into a single dot since you can move from $r_{2i-1}$ to $r_{2i}$ by moving up once then left once, and similarly for the triangle on the left and the one extra dot in row $4$.
A different way to prove this is to check $n=3$, then note that the movement allowed partitions the triangulation of the plane into four sets of triangles.(in a triangle of height $2$ with four points, the points can visit every triangle in the plane, but no two points can visit the same triangle.) Then noting that $n^2/4\geq n$ for each all $n\geq 4$ gives us that one of the four sets in the partition must intersect our chosen triangle of height $n$ in at least $n$ spots. 
