Claim 1: If $n$ is divisible by $3$, then $n$ doesn't satify Proof We color all the vertices consecutively with $3$ colors red, blue and yellow such that every equilateral triangle has exactly $3$ different colors Then since $n$ is divisible by $3$ then the number of red vertices is $1$ more than the number of blue vertices and the yellow vertices Let the number of red, blue and yellow vertices which are tail are $A,B,C$ respectively Consider $A-B,B-C,C-A$ Since every move, the number of a type of vertice which is tail is both increase or decrease by $1$ which means $A-B,B-C,C-A$ are always in the same parity (odd or even) Since at the beginning $A-B,B-C,C-A$ are all even ( $=0$) but the number of red vertices is $1$ more than the number of blue vertices and the yellow vertices, which gives a contradiction! Claim 2: If $n$ isn't divisible by $3$, then $n$ satifies Proof We construct the configuration satisfies for every $n$ isn't divisible by $3$ The easiest case is the the case $n=1$ and $n=2$ The case $n=1$ is trivial. With the case $n=2$, just flip the coin in the order below and we get all coins tail up For $n \equiv 1 \pmod 3$ We prove by induction. Suppose for $n-3$ we can get all coins tail up, then for the last $3$ lines, we flip the coin by the group of $n=2$ (I mean the group of equilateral triangle with the length $2$) and the group of $n=1$ as colored below The particular case of the last $3$ lines when $n=7$ make it more obvious For $n \equiv 2 \pmod 3$ We prove by induction. Suppose for $n-3$ we can get all coins tail up, then for the last $3$ lines, we flip the coin by the group of $n=2$ and the group of $n=1$ as colored below Thus the answer is $n$ that isn't divisible by $3$