Let the triangle be partitioned into $N^2$ equal equilateral triangles, with $N=2n-1$, $n\geq 3$, labelled by the numbers $1$ to $N^2$; and prove there exist two adjacent such triangles with labels differing by more than $\dfrac {N+1}{2} = n$ (the problem at hand is for $n=3$). Consider the graph with vertices the $N^2$ triangles, and edges connecting the centres of adjacent triangles (it is embedded in a hexagonal lattice). For two vertices $u,v$, define as usual $\operatorname{d}(u,v)$ to be the length of the shortest path connecting $u$ and $v$. It is immediate that $\operatorname{d}(u,v) \leq 2N-2 = 4(n-1)$, with equality only if one of $u,v$ is one corner triangle, and the other is adjacent to the opposite side of the large triangle. Now, if the label of $u$ belongs to $\{1,2\}$ and the label of $v$ belongs to $\{N^2-1,N^2\}$, if $\operatorname{d}(u,v) \leq 2N-3$ then, since $1 + n(2N-2) = 1 + 4n(n-1) = (2n-1)^2 = N^2$, it follows that $2+n(2N-3) < N^2-1$, so at least one difference of consecutive labels along the path must be larger than $n$. On the other hand, if $\operatorname{d}(u,v) = 2N-2$, it follows that each of the vertices labelled $\{1,2,N^2-1,N^2\}$ must be a corner triangle, impossible since there are only three corner triangles.