Problem 1829

Solution 1. Write $a = dr$ and $b = ds$ where $d = \gcd(a, b)$ and $(r, s) = 1$. Then $d^{2021} r^{2021} s^{2021}$ divides $n$. Furthermore $2 \cdot 3 \cdot 337 \mid d(r - s)$.

Case 1: Assume $337 \mid d$. Since $(r, s) = 1$, we may assume that $r$ is odd. Then
$$\{337r^2, 337r^4, \dots, 337r^{10}, 337r^{12}\}$$
works. Indeed each of these six divisors of $n$ is congruent to $1 \mod 4$ so their sum is a multiple of $2$ but not of $4$. If $3 \mid r$ then the sum is $0 \mod 3$ while if $3 \nmid r$ then each of these divisors is $1 \mod 3$ so the sum is again $0 \mod 3$. Therefore the sum is a multiple of $2022$ but not of $2022^2$.

Case 2: Assume $337 \nmid d$. Then $337 \mid r - s$ and since $(r, s) = 1$ then $337 \nmid rs$. Consider the $2022^2$ divisors of $n$ of the form $r^k s^\ell$ where $k, \ell \in \{0, 1, 2, \dots, 2021\}$. Since none of them is a multiple of $337$, they have at most $2 \cdot 3 \cdot 336$ distinct remainders modulo $2022$. Therefore at least $\frac{2022^2}{2 \cdot 3 \cdot 336} > 2022$ of them have the same remainder modulo $2022$.
Pick $2023$ out of those, say $d_1, d_2, \dots, d_{2023}$. Let $S$ be their sum. We claim that there is a subset of $2022$ of them that will work. Note that the sum of any such subset is a multiple of $2022$. It is enough to show that there is such a subset whose sum is not divisible by $337^2$. If this is not the case then $S - d_i \equiv 0 \mod 337^2$ for each $i = 1, 2, \dots, 2023$. In particular, all $d_i$ are congruent mod $337^2$. Say that $d_i \equiv k \mod 337^2$ for each $i$. Then $337 \nmid k$ and so the sum of any $2022$ of them is congruent to $2022k \neq 0 \mod 337^2$.

We start with the following claim:
Claim. If $k$ is a positive integer, then $a^k b^{2021-k} \mid n$.
Proof of the Claim. We have that $n^{2021} = n^k \cdot n^{2021-k}$ is divisible by $a^{2021k} \cdot b^{2021(2021-k)}$ and taking the $2021$-root we get the desired result. $\square$

Back to the problem, we will prove that the set $T = \{a^k b^{2021-k}, k \ge 0\}$ consisting of $2022$ divisors of $n$, has the desired property. The sum of its elements is equal to
$$S = \sum_{k=0}^{2021} a^k b^{2021-k} \equiv \sum_{k=0}^{2021} a^{2021} \equiv 0 \mod 2022.$$
On the other hand, the last sum is equal to $\frac{a^{2022} - b^{2022}}{a-b}$. We will prove that this is not divisible by $9$. Indeed, if $3^t \mid a - b$ then, since $3^1 \mid 2022$, by the Lifting the Exponent Lemma, we have that $3^{t+1} \mid a^{2022} - b^{2022}$. This implies that $S$ is not divisible by $9$, therefore, $2022^2$ doesn't divide $S$.