Romanian Mathematical Olympiad - Final Round

We notice that $a > b$, for all $a \in A$ and $b \in B$. Indeed, the contrary would imply that $a < b$, so the remainder of the division of $a$ by $b$ is $a$ and $a$ belongs to both $B$ and $C$. This shows that $A$ is made by consecutive numbers and $2n+1 \in A$.

Take $c \in C$. From the assumption there exists $a \in A$ and $b \in B$ so that $a = bq+c$. Then $a \ge (c+1) \cdot 1 + c = 2c + 1$, hence $2n+1 \ge 2c+1$, therefore $n \ge c$. If we suppose that $n+1 \notin B$, then $n+1 \in A$, whence $\{n+1, n+2, \dots, 2n+1\} \subset A$. So, if $b \in B$, then $b \le n$ and, since $A$ contains $n+1$ consecutive numbers, $A$ must contain at least one multiple of $b$ – false.

Henceforth $n+1 \in B$, so there exists $k \in \{n+1, n+2, \dots, 2n\}$ so that $A = \{2n+1, 2n, \dots, k+1\}$. From $c \le n$ for all $c \in C$ follows $\{n+1, n+2, \dots, k\} \subset B$.

The remainders of the divisions of the elements of $A$ to the elements of $\{n+1, n+2, \dots, k\}$ are the numbers $\{1, 2, \dots, n\}$. This shows that $B = \{n+1, n+2, \dots, k\}$ and $C = \{1, 2, \dots, n\}$. The number $k$ can be every element of the set $\{n+1, n+2, \dots, 2n\}$, so there are $n$ possible splits.