TITLE: Partition of $\{1,2,3,\cdots,3n\}$ into $n$ subsets, each with $3$ numbers, which have equal sum
QUESTION [5 upvotes]: I want to show, that for every odd $n$ $(n\ge3)$, there exists a partition of $\{1,2,3,\cdots,3n\}$ into $n$ disjoint subsets, where each one has $3$ elements and equal sum.
The first such number is $3$. For $3$ it is obvious. $\{1,6,8\}, \{2,4,9\}, \{3,5,7\}$. I tried to show this using induction, but it seems I have some trouble with it. Please help me, if you can.

REPLY [5 votes]: We let $k$ range from $1$ to $n$.  Our sets are $$\begin {cases}
\{k, \frac{3n-1}2+k,3n+2-2k\} &1\le k \le \frac {n+1}2\\
\{k,n+k-\frac{n+1}2,4n-2k+2\}&\frac{n+1}2 \lt k \le n
\end {cases}$$
These can be seen to add to $\frac {9n+3}2$ and to use the numbers $1$ to $n$ in the first entry, $n+1$ to $2n$ in the second and $2n+1$ to $3n$ in the third.  They follow the pattern in Ng Chung Tak's answer for $n=5$