Taiwan IMO Selection Camp, 2023

Easy to prove that $P$ must have rational coefficients.
If $P$ has a degree $d \ge 2$, consider $P(x) = \frac{p}{q}x^d + \frac{1}{M}Q(x)$, where $p, q \in \mathbb{Z}$, $Q(x)$ is a polynomial of degree $< d$ with integer coefficients. Suppose $P\left(\frac{m}{n}\right) = \frac{1}{n}$, $m, n$ coprime.
We have $Mn^{d-1}P\left(\frac{m}{n}\right) \in \mathbb{N}$, thus $qn \mid pMm^d$, and so $n \mid pM$. However, $n$ can be arbitrarily large, so $pM = 0$, contradiction. Therefore, we must have $d = 1$.
Assume $P(x) = \frac{p x}{q} + \frac{r}{s}$, $\gcd(p, q) = \gcd(r, s) = 1$. The condition is equivalent to that for infinitely many coprime positive integer pairs $(m, n)$,
$$psm + rqn = qs.$$
Let $A = ps$, $B = qr$, $C = qs$. We have $\gcd(A, B) \mid C$. We also must have $AB \le 0$, otherwise $|C| = |Am + Bn| \ge |m + n|$, contradicting the existence of infinitely many solutions.
Conversely, if $\gcd(A, B) \mid C$ and $AB \le 0$, we can find integers $a, b$ such that $Aa + Bb = \gcd(A, B)$. Let $m = |B|t + Ca$, $n = |A|t + Cb$, we have $Am + Bn = C$, and $\gcd(m, n) \mid \gcd(C, am - bn) = \gcd(C, t + s)$, where $s$ is some integer independent of $t$. Therefore, we have found large enough $t$ so that $m, n$ are positive and coprime.
So all solutions of $P(x)$ are $P(x) = \frac{p x}{q} + \frac{r}{s}$, such that $\gcd(ps, qr) \mid qs$ and $\frac{pr}{qs} \le 0$. Since $\gcd(p, q) = \gcd(r, s) = 1$, so $\gcd(ps, qr) = \gcd(p, r)\gcd(q, s)$, the first condition is equivalent to $\gcd(p, r) = 1$.