Find all $n$ such that $\{1,2,\dots,n\}$ be divided into disjoint triples such that for each triple, one element is the sum of the other two elements.
On a 2013 combinatorics handout (high school Math Olympics) of tian27546西西 (translated into English)
Assume $\{1,2,\dots,n\}$ be divided into disjoint triples, then $3\mid n$. And because each triple can be written in the form of $(a,b,a+b)$, the sum of the elements of each triple is even, so $1+2+\dots+n=n(n+1)/2$ is even, so $n$ must be in the form of $12k$, or $12k+3$.
Lemma: Assume that there is a division for $n=k$, then there is a division for $n=4k$ or $n=4k+3$.
Proof: $\{2,4,⋯,2k\}$ can be divided in the same way as $\{1,2,⋯,n\}$. For $n=4k$, the remaining numbers $\{1,3,⋯,2k-1,2k+1,2k+2,⋯,4k-1,4k\}$ can be divided into $k$ triples $(2j-1,3k-j+1,3k+j),j=1,2,⋯,k$, as follows (each column is a triple): \begin{pmatrix}1 & 3 & 5 & \cdots & 2 k-3 & 2 k-1 \cr 3 k & 3 k-1 & 3 k-2 & \cdots & 2 k+2 & 2 k+1 \cr 3 k+1 & 3 k+2 & 3 k+3 & \cdots & 4 k-1 & 4 k\end{pmatrix} For $n=4k+3$, the remaining numbers $\{1,3,⋯,2k-1,2k+1,⋯,4k+2,4k+3\}$ can be divided into $k+1$ triples $(2i-1,3k+3-j,3k+j+2),j=1,2,⋯,k+1$, we also get the following (each column is a triple) :\begin{pmatrix}1 & 3 & 5 & \cdots & 2 k-1 & 2 k+1 \cr 3 k+2 & 3 k+1 & 3 k & \cdots & 2 k+3 & 2 k+2 \cr 3 k+3 & 3 k+4 & 3 k+5 & \cdots & 4 k+2 & 4 k+3\end{pmatrix} Example: Since for $n=3$ there is a division, and$$3 \longrightarrow_{3 \times 4+3} 15 \longrightarrow_{15 \times 4} 60 \longrightarrow_{60 \times 4+3} 243 \longrightarrow_{243 \times 4+3} 975$$so there is a division for $n=975$.
All $n$ that can be reduced to $3$, are $3, 12, 15, 48, 51, 60, 63,\dots$(See A001196)
There are $n$ that satisfies the condition but can not be reduced to $3$, the smallest ones are $24,27,36,39$. Using the lemma, each of them can produce other numbers. \begin{array}l n=24& \begin{pmatrix} 1 & 3 & 5 & 8 & 4 & 7 & 6 & 2 \\ 23 & 19 & 16 & 12 & 14 & 10 & 9 & 11 \\ 24 & 22 & 21 & 20 & 18 & 17 & 15 & 13 \end{pmatrix}\\ n=27& \begin{pmatrix} 5 & 3 & 6 & 9 & 8 & 7 & 4 & 2 & 1 \\ 11 & 14 & 12 & 10 & 13 & 15 & 20 & 23 & 26 \\ 16 & 17 & 18 & 19 & 21 & 22 & 24 & 25 & 27 \end{pmatrix}\\ n=36&\begin{pmatrix} 6 & 8 & 4 & 7 & 12 & 11 & 10 & 9 & 5 & 3 & 2 & 1 \\ 15 & 14 & 19 & 17 & 13 & 16 & 18 & 20 & 26 & 30 & 32 & 35 \\ 21 & 22 & 23 & 24 & 25 & 27 & 28 & 29 & 31 & 33 & 34 & 36 \end{pmatrix}\\ n=39&\begin{pmatrix} 7 & 5 & 8 & 6 & 13 & 12 & 11 & 10 & 9 & 4 & 3 & 2 & 1 \\ 15 & 19 & 17 & 20 & 14 & 16 & 18 & 21 & 23 & 30 & 33 & 35 & 38 \\ 22 & 24 & 25 & 26 & 27 & 28 & 29 & 31 & 32 & 34 & 36 & 37 & 39 \end{pmatrix} \end{array} Originally posted by user realnumber on Math Entertainment forum. I found the above divisions for $24,27,36,39$ by Mathematica.
A108235 Number of partitions of $\{1,2,...,3n\}$ into $n$ triples $(X,Y,Z)$ each satisfying $X+Y=Z$. $$\{a_1,a_2,\dots\}=\{1, 0, 0, 8, 21, 0, 0, 3040, 20505, 0, 0, 10567748, 103372655,\\ 0, 0, 142664107305, 1836652173363, 0, 0,\dots\}$$ The desired sequence is$$\{3n:a_n\ne0\}=\{3, 12, 15, 24, 27, 36, 39, 48, 51,\dots\}$$