IMO Team Selection Test

Call a positive integer $n$ 'good' if there exists $m$ that $m$ in the interval $\left(\frac{2n}{3}, n\right)$ and $\frac{m}{n-m}, \frac{n-m}{m}$ is not pretty. Next, we will prove the following claims.

Claim 1. If $n$ is good then any multiple of $n$ is also good.

Proof. Consider a good number $n$, there exists $m \in \left(\frac{2n}{3}, n\right)$ such that $\frac{n-m}{m}$ and $\frac{m}{n-m}$ are not pretty. Hence, $kn$ is not a good number because $\frac{kn-km}{km}, \frac{km}{kn-km}$ are not pretty. $\square$

Claim 2. Consider a prime $q$ such that there exists a prime $r$ that $2022 < r < q$. For all pairs $(p, k)$ with $p$ is prime and $k$ is positive integer such that $p^k > 3 \cdot q!$, number $p^k$ is good.

Proof. Choose $m = 3 \cdot q!$ then $\frac{2p^k}{3} < p^k - m < p^k$. Assume that $\frac{p^k - m}{m}$ is pretty then there exists $b < 2023$ and finite non-negative integers $b_0, b_1, b_2, \dots, b_t$ that
$$\frac{p^k - m}{m} = \sum_{i=0}^{t} \frac{b_i}{b^i} = \frac{b_0 b^t + b_1 b^{t-1} + \dots + b_t}{b^t},$$
which means $\frac{m}{\text{gcd}(m, p^k)} \mid b^t$.
Hence, all prime divisors of $m$, except $p$, are smaller than $2022$. On the other hand, $m$ is divisible for $q, r > 2022$ which implies that $m$ still has a prime divisor larger than $2022$ and differs from $2022$, which is a contradiction. Therefore, $\frac{p^k - m}{m}$ is not pretty.
Continue, assume that $\frac{m}{p^k - m}$ is pretty. Similarly, $\frac{p^k - m}{\text{gcd}(m, p^k - m)}$ only has prime divisors that are smaller than $2022$. Assume that there exists a prime number $p_1 < 2022$ such that
$$p_1 \mid \frac{p^k - m}{m} \longrightarrow p_1 \mid p^k \longrightarrow p_1 = p,$$
which means $p < 2022$ and there exists $l > 0$ that
$$p^l \cdot \text{gcd}(m, p^k - m) = p^k - m.$$
Let $m = p^s \cdot t$ where $(p, t) = 1$, the above equation becomes
$$p^l \cdot p^s = p^k - p^s \cdot t \text{ or } p^{k-s} - t = p^l.$$
Note that $p^{k-s} > 1$ then $p \mid p^{k-s}$. We also have $\text{gcd}(t, p) = 1$ then $\text{gcd}(p^l, p) = 1$ or $l = 0$, which means $\frac{p^k - m}{\text{gcd}(m, p^k - m)} = 1$ or $p^k - m \mid m < \frac{p^k}{3}$, which is a contradiction.
Hence, both $\frac{m}{p^k - m}$ and $\frac{p^k - m}{m}$ are not pretty, or $p^k$ is good. $\square$

Back to the original problem, let $N$ be the number of the numbers that have the form $p^k$ and not exceed $Q = 3 \cdot q!$. We will point out that all positive integers $n > Q^N$ is good. Assume that the prime factorization of $n$ is as follows
$$n = \prod_{i=1}^{t} p_i^{\alpha_i}.$$
If $p_i^{\alpha_i} < Q$ for all $i \le t$ then $n \le Q^t \le Q^N$, a contradiction.
Therefore, there exists an index $i$ such that $p_i^{\alpha_i} > Q$, which means $p_i^{\alpha_i}$ is good by Claim 2. Hence, $n$ is the multiple of $p_i^{\alpha_i}$ and $n$ is good by Claim 1. The problem is solved. $\square$