Partition of $\{1,2,3,\cdots,3n\}$ into $n$ subsets, each with $3$ numbers, which have equal sum 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.
 A: 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 $n=5$
A: Let $S$ be the equal sum of each partition, then $nS=1+\ldots+3n$.
i.e. $\: nS=\frac{3n(3n+1)}{2} \implies S=\frac{3(3n+1)}{2}$
Now, $n=1$ is trivial and $n$ should be odd since the sum is an integer.
Note that the median of the sequence is $\frac{3n+1}{2}$.
Observing $1+\frac{3n+1}{2}+3n=S$
Take $a+b+c=0$:
$(1+a)+(\frac{3n+1}{2}+b)+(3n+c)=S$
By trial and error, one possible set of $n=5$ is
$$
\begin{array}{|c|c|c|c|c|c|}
\hline
a & 0 &  1 &  2 &  3 &  4 \\
\hline
b & 0 &  1 &  2 & -2 & -1 \\
\hline
c & 0 & -2 & -4 & -1 & -3 \\
\hline
\end{array}
$$
That is,
$$
\begin{array}{|c|c|c|c|c|c|}
\hline
p & 1 & 2 & 3 & 4 & 5 \\
\hline
q & 8 & 9 & 10 & 6 & 7 \\
\hline
r & 15 & 13 & 11 & 14 & 12 \\
\hline
\end{array}
$$
Also see the links of similar question and magic rectangle
Snapshots one, two and three from Thomas R. Hagedorn, Magic rectangles revisited, Discrete Mathematics 207 (1999), 65-72.
