For which $n\in \mathbb{N}$ can we divide the set $\{1,2,3,\ldots,3n\}$ into $n$ subsets each with $3$ elements such that in each subset $\{x,y,z\}$ we have $x+y=3z$?

Since $x_i+y_i=3z_i$ for each subset $A_i=\{x_i,y_i,z_i\}$, we have $$4\sum _{i=1}^n z_i=\sum _{i=1}^{3n}i = {3n(3n+1)\over 2} \implies 8\mid n(3n+1) $$ so $n=8k$ or $n=8k-3$. Now it is not difficult to see that if $k=1$ we have such partition.

- For $n=5$ we have: $$A_1= \{9,12,15\}, A_2= \{4,6,14\}, A_3= \{2,5,13\}, \\A_4= \{10,7,11\}, A_5= \{1,3,8\}$$
- For $n=8$ we have: $$A_1= \{24,21,15\}, A_2= \{23,19,14\}, A_3= \{22,2,8\}, A_4= \{20,1,7\}, \\A_5= \{17,16,11\}, A_6= \{18,12,10\}, A_7= \{13,5,6\}, A_8= \{9,3,4\}$$

What about for $k\geq 2$? Some clever induction step? Or some ''well'' known configuration?

Source: Serbia 1983, municipal round, 3. grade