From the possibility of partitioning the set into disjoint triples it follows that $3 \mid n$. In each triple $\{a, b, a+b\}$ the sum of its elements is $2(a+b)$, hence an even number; thus also the sum of all numbers from $1$ to $n$ must be even, i.e. the product $n(n+1)$ must be divisible by four. Altogether it therefore follows that the number $n$ has to be of the form either $12k$ or $12k+3$; from the given set of numbers, this is satisfied only for $n = 3900$ and $n = 3903$.

In the next paragraph we describe a construction how to produce, starting from a decomposition satisfying the given condition for some $n = k$, a decomposition of the same kind for $n = 4k$ and $n = 4k + 3$. This guarantees that the required decompositions for $n = 3900$ and $n = 3903$ indeed exist, in view of the decreasing sequence
$$3900 \rightarrow 975 \rightarrow 243 \rightarrow 60 \rightarrow 15 \rightarrow 3$$
(instead of $3900$ one can start also with $3903$) and the trivial decomposition for $n=3$ (from which we in turn construct the decompositions for $n=15$, $n=60$ etc. up to $n=3900$ or $n=3903$).

From a decomposition of the set $\{1, 2, \dots, k\}$ satisfying the given conditions we first produce a similar decomposition for the set of the first $k$ even numbers $\{2, 4, \dots, 2k\}$ (simply by multiplying all the numbers in the triples by two). In the case of $n=4k$ we partition the remaining numbers
$$\{1, 3, 5, \dots, 2k-1, 2k+1, 2k+2, \dots, 4k-1, 4k\}$$
into the $k$ triples $\{2j-1, 3k-j+1, 3k+j\}$, where $j=1, 2, \dots, k$. They are shown in the columns of the table below.

$$\begin{pmatrix} 1 & 3 & 5 & \dots & 2k-3 & 2k-1 \\ 3k & 3k-1 & 3k-2 & \dots & 2k+2 & 2k+1 \\ 3k+1 & 3k+2 & 3k+3 & \dots & 4k-1 & 4k \end{pmatrix}$$

In the case of $n = 4k + 3$ we partition the remaining numbers
$$\{1, 3, 5, \dots, 2k-1, 2k+1, 2k+2, \dots, 4k+2, 4k+3\}$$
into the $k+1$ triples $\{2j-1, 3k+3-j, 3k+j+2\}$, where $j=1, 2, \dots, k+1$; these are again shown in the columns of the table below.

$$\begin{pmatrix} 1 & 3 & 5 & \dots & 2k-1 & 2k+1 \\ 3k+2 & 3k+1 & 3k & \dots & 2k+3 & 2k+2 \\ 3k+3 & 3k+4 & 3k+5 & \dots & 4k+2 & 4k+3 \end{pmatrix}$$

This completes the proof of the fact that the solution of the given problem are the numbers $n=3900$ and $n=3903$.