CGMO, 2022

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
$$A' = \{a+xn \mid a \in A, x = 0, 1, \dots, d-1\}, \quad B' = \{b+xn \mid b \in B, x = 0, 1, \dots, d-1\}$$
We verify that $A', B'$ satisfy the problem requirements for $dn$.
Since the remainders of $A', B', A' + B'$ modulo $n$ are exactly the remainders of $A, B, A+B$ modulo $n$, the remainders of $A', B', A' + B'$ 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 $m \equiv a \pmod{n}$, 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 $m \equiv b \pmod{n}$, then (ii) holds. If there exist $a \in A, b \in B$, such that $m \equiv a+b \pmod{n}$, 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. $A', B'$ 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. $\Box$