Czech-Slovak-Polish Match

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 $$3900\to 975\to 243\to 60\to 15\to 3$$ (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.

$$\left(\begin{array}{cccccc}1& 3& 5& \dots & 2k-3& 2k-1\\ 3k& 3k-1& 3k-2& \dots & 2k+2& 2k+1\\ 3k+1& 3k+2& 3k+3& \dots & 4k-1& 4k\end{array}\right)$$

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.

$$\left(\begin{array}{cccccc}1& 3& 5& \dots & 2k-1& 2k+1\\ 3k+2& 3k+1& 3k& \dots & 2k+3& 2k+2\\ 3k+3& 3k+4& 3k+5& \dots & 4k+2& 4k+3\end{array}\right)$$

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