2022 CGMO

Proof. Let $A+B=\{a+b\mid a\in A,b\in B\}$. The problem can be understood as the remainders of $A$ modulo $n$, the remainders of $B$ modulo $n$, and the remainders of $(A+B)$ modulo $n$ form a partition of all remainders modulo $n$.

(1) If $n>1$ is an odd number, let $n=2k+1$, $k\in {\mathbb{Z}}_{>0}$. Take $A=\{k\}$, $B=\{k+1,k+2,\dots ,2k\}$, then $A+B=\{2k+1,2k+2,\dots ,3k\}$, so $A,B,A+B$ are pairwise disjoint, and their union is exactly the consecutive $2k+1$ numbers, forming a complete system modulo $2k+1$, satisfying the condition.

(2) If $n>1$ satisfies the condition, then for any integer $d>1$, $dn$ also satisfies the condition. In fact, let the sets corresponding to $n$ be $A,B$. Set $$A^{\prime }=\{a+xn\mid a\in A,x=0,1,\dots ,d-1\},B^{\prime }=\{b+xn\mid b\in B,x=0,1,\dots ,d-1\}$$ We verify that $A^{\prime },B^{\prime }$ satisfy the problem requirements for $dn$. Since the remainders of $A^{\prime },B^{\prime },A^{\prime }+B^{\prime }$ modulo $n$ are exactly the remainders of $A,B,A+B$ modulo $n$, the remainders of $A^{\prime },B^{\prime },A^{\prime }+B^{\prime }$ modulo $n$ and modulo $dn$ have no intersection, so no more than one statements of (i), (ii), and (iii) can hold. Considering any integer $m$, if there exists $a\in A$ such that $m\equiv a(\mathrm{mod}n)$, then $m$ is congruent modulo $dn$ with one of $a+xn$ ($x=0,1,\dots ,d-1$), (i) holds. Similarly, if there exists $b\in B$ such that $m\equiv b(\mathrm{mod}n)$, then (ii) holds. If there exist $a\in A,b\in B$, such that $m\equiv a+b(\mathrm{mod}n)$, then $m$ is congruent modulo $dn$ with one of $a+(b+xn)$ ($x=0,1,\dots ,d-1$), (iii) holds. Therefore, exactly one of (i), (ii), and (iii) holds. $A^{\prime },B^{\prime }$ satisfy the problem requirements for $dn$. By (1) and (2), all integers $n>1$ except powers of 2 satisfy the problem conditions.

(3) When $n=8$, it is easy to verify that $A=\{1,2\}$, $B=\{3,6\}$, $A+B=\{4,5,7,8\}$, satisfying the requirements. Combining with (2), we know that all $2^k$ ($k\ge 3$) satisfy the requirements.

Since $A,B,A+B$ are all non-empty sets, there are at least three different remainders modulo $n$, $n\ge 3$, so $n=1,2$ do not satisfy the conditions. When $n=4$, if $A,B$ have only one remainder modulo 4, then $A+B$ also has only one remainder modulo 4, which does not satisfy the requirement. If either $A$ or $B$ has at least two different remainders modulo 4, assume $a_1,a_2\in A$ have different remainders modulo 4, and take any $b\in B$, then $a_1+b$ and $a_2+b$ also have different remainders modulo 4, so $A+B$ has at least two remainders modulo 4, which also does not satisfy the requirement. Therefore, $n=4$ does not satisfy the condition either.

In conclusion, the desired $n$ are all positive integers except 1, 2, and 4. $\square$