If $P(x)$ is constant, then we must have $P(x)\equiv \frac 1k$ for some positive integer $k$. Assume that $deg(P)\ge 1$. Clearly, there are infinitely many rational numbers $r$ such that $P(r)\in\mathbb{Q}$. Then, by the Lagrange Interpolation formula, we know that $P$ has rational coefficients. Let $P(x)=q_tx^t+q_{t-1}x^{t-1}+\cdots +q_1x+q_0$ where $q_i\in\mathbb{Q}$ and $t\in\mathbb{Z^+}$. Let $A$ be the smallest positive integer such that for all $0\leq i\leq t$, we have $Aq_i\in\mathbb{Z}$. Suppose that $deg(P)=1$. Let $P(x)=\frac abx+\frac cd$ where $a,b,c,d\in\mathbb{Z}\; ;\; b,d>0$ and $(a,b)=(c,d)=1$. If $c=0$, then, $P\left(\frac mn\right)=\frac 1n\Leftrightarrow b=am.$ Hence, we must have $a=1$ (Because as $(a,b)=1$, we get $|a|=1$. If $a=-1$, then we must have $m=-b<0$, which contradicts the problem statement). Also, if $c=0, a=1$, we can choose $m=b$ and $n$ can be any integer coprime with $b$. Thus, $P(x)=\frac xk$ is a solution where $k\in\mathbb{Z^+}$. Suppose that $c\neq 0$. We have $P\left(\frac mn\right)=\frac 1n\Leftrightarrow adm+cbn=bd$. Assume that there exists a prime number $p$ such that $p|(a,c)$. Then, we must have $(p,bd)=1$. But, we have $p|adm+cbn=bd$, contradiction. Hence, $a$ and $c$ must be coprime. Also, if $a,c>0$, then clearly the equation $adm+cbn=bd$ has finitely many solutions over positive integers. If $a,c<0$, the equation has $0$ solutions over positive integers. Hence, we have $ac<0$. Now, let's show that if $(a,c)=1$ and $ac<0$, then there exist infinitely many coprime positive integers $m,n$ such that $adm+cbn=bd$. Let $e=(b,d),\; b=eb_1,\; d=ed_1$. Let $m=b_1m_1, n=d_1n_1$. We have $adm+cbn=bd\Leftrightarrow am_1+cn_1=e$. As $(a,c)=1$, we know that this equation has a solution over integers. Let $au+cv=e$ where $u,v\in\mathbb{Z}$. Now, by CRT, the system of equations $y\equiv \frac{1-u}{c}\pmod{\frac {ed_1}{(e,b_1)}}$ and $y\equiv \frac{v-1}{a}\pmod{b_1}$ have common solutions. Let $k\in\mathbb{Z}$ be any of them. These equations implies $(ck+u, d_1)=(-ak+v, b_1)=1$. See that if we let $m_1=ck+u,\; n_1=-ak+v$, then $am_1+cn_1=e$. Assume that there exists a prime number $s$ such that $s|m_1, n_1$. Then, $s|e$. As $(n_1, b_1)=1$, we get $(s,b_1)=1$. Also, we have $m_1=ck+u\equiv 1\pmod{\frac {ed_1}{(e,b_1)}}.$ As $s|\frac {ed_1}{(e,b_1)}$, we get a contradiction. Therefore, $(m_1,n_1)=1$. Overall, we find that $m=b_1m_1, n=d_1n_1$ and $(n_1,b_1)=(m_1,d_1)=(m_1,n_1)=(b_1,d_1)=1$. Hence, $(m,n)=1$. Now, if $c>0, a<0$, then for all sufficiently large integers $k$, we can choose $m=b_1(ck+u), n=d_1(-ak+v)$ and we will ensure that $m,n>0$. Hence, there will be infinitely many coprime integers $m,n$ such that $P\left(\frac mn\right)=\frac 1n$. If $c<0, a>0$, then, similarly, for all sufficiently small integers $k$, we can choose $m=b_1(ck+u), n=d_1(-ak+v)$ and we will get $m,n>0$. Suppose that $deg(P)\ge 2.$ Let $P\left(\frac mn\right)=\frac 1n$. Then, $Aq_t\left(\frac mn\right)^t+\cdots +Aq_1\left(\frac mn\right)+Aq_0=\frac An$. By multiplying both sides with $n^{t-1}$, one can find that $Aq_t\left(\frac{m^t}n\right)\in\mathbb{Z}$. Hence, $n|Aq_t$. As $Aq_t\neq 0$, we find that $n$ can take a finite number of values. Thus, there exists a positive integer $n_0$ for which there exists infinitely many positive integers $m$ such that $(n_0,m)=1$ and $P\left(\frac m{n_0}\right)=\frac 1{n_0}$. Define $Q(x)=\frac{q_t}{n^t}x^t+\frac{q_{t-1}}{n^{t-1}}x^{t-1}+\cdots +\frac{q_1}{n}x+q_0$. See that the condition $P\left(\frac m{n_0}\right)=\frac 1{n_0}$ implies $Q(m)=\frac 1{n_0}$. $Q$ takes the same value infinitely many times, hence $Q$ must be constant. Yields, $t=0$, and this gives us a contradiction. @below, thanks for letting me know. I totally missed the case $t=1$. Hope this works.

BarisKoyuncu wrote: If $P(x)$ is constant, then we must have $P(x)\equiv \frac 1k$ for some positive integer $k$. Assume that $deg(P)\ge 1$. Clearly, there are infinitely many rational numbers $r$ such that $P(r)\in\mathbb{Q}$. Then, by the Lagrange Interpolation formula, we know that $P$ has rational coefficients. Let $P(x)=q_tx^t+q_{t-1}x^{t-1}+\cdots +q_1x+q_0$ where $q_i\in\mathbb{Q}$ and $t\in\mathbb{Z^+}$. Let $A$ be the smallest positive integer such that for all $0\leq i\leq t$, we have $Aq_i\in\mathbb{Z}$. Let $P\left(\frac mn\right)=\frac 1n$. Then, $Aq_t\left(\frac mn\right)^t+\cdots +Aq_1\left(\frac mn\right)+Aq_0=\frac An$. By multiplying both sides with $n^{t-1}$, one can find that $Aq_t\left(\frac{m^t}n\right)\in\mathbb{Z}$. Hence, $n|Aq_t$. As $Aq_t\neq 0$, we find that $n$ can take a finite number of values. Thus, there exists a positive integer $n_0$ for which there exists infinitely many positive integers $m$ such that $(n_0,m)=1$ and $P\left(\frac m{n_0}\right)=\frac 1{n_0}$. Define $Q(x)=\frac{q_t}{n^t}x^t+\frac{q_{t-1}}{n^t}x^{t-1}+\cdots +\frac{q_1}{n^t}x+\frac{q_0}{n^t}$. See that the condition $P\left(\frac m{n_0}\right)=\frac 1{n_0}$ implies $Q(m)=\frac 1{n_0}$. $Q$ takes the same value infinitely many times, hence $Q$ must be constant. Yields, $t=0$, and this gives us a contradiction. I think you miss a (large) family of solutions. EDIT: @above the solution (at the large scale) seems fine

The answer is $P(x)=ax+b$ where there doesn’t exist a prime $p$ such that $\nu_p(a), \nu_p(b)$ are both positive, or both negative. By Lagrange Interpolation, $P(x) \in \mathbb{Q}[x]$. Case 1: $\deg P\ge 2$. If $\deg P\ge 2$, the key intuition is that there is a large prime power dividing $n$, call $p^c$. Let $P(x) = \sum_{j=0}^d a_jx^j$ and $p=\max\{ q \mid q \text{ prime } \nu_q(a_j)\ne 0 \text{ for some j }\}$ and suppose that $|\nu_q(a_k)| \le M$ for all $0\le k\le d$. Let $K $ be a large positive integer such that all integers greater than $K$ have a prime power of size at least $p^{3M}$. Then when $n>K$ say $p^{3M} < q^c | n$, then we can show with size that $\nu_q(a_d (m/n)^d) < \nu_q(a_k (m/n)^k)$ for all $k=0,\cdots,d-1$ so $\nu_q(P(m/n)) = \nu_q(a_d (m/n)^d) < \nu_q(1/n)$. This implies $n$ is bounded. For each $n\le K$ the number of $m$ must be finite, so there are finitely many coprime positives $(m,n)$ such that $$P\left( \frac mn \right)=\frac 1n $$ If $\deg P\le 1$ then let $P(x)=ax+b$ where $a,b\in \mathbb{Q}$. Then $P(m/n)=am/n+b=1/n$ has infinitely many coprime solutions $(m,n)$. In other words $am+bn=1$. For $\nu_p$ reasons $\nu_p(a),\nu_p(b)$ cannot be both negative for the same $p$ or positive for the same $p$. Let $S_a^+ = \{ p | \nu_p(a)>0\}$. Define $S_a^-, S_b^+, S_b^-$ similarly. We already know that $S_a^+ \cap S_b^+ = S_a^- \cap S_b^- =\emptyset$. Let $a’,b’$ be smallest possible integers such that $a’a, b’b\in \mathbb{Z}$. Then $aa’c+bb’d=1$ where $m=a’c, n=b’d$. We know that $a’$ only have prime factors in $S_a^-$ and $b’$ only have prime factors in $S_b^-$. Note $\nu_p(aa’) >0$ only if $\nu_p(a) > 0 \iff p\in S_a^+$, so $aa’, bb’$ are coprime integers. We need to guarantee $\gcd(a’c,b’d)=1$. For brevity let $x=aa’, y=bb’$. Then I have $xc+yd=1$ and the solutions $(c,d)$ can be of the form $(c_0+yk, d_0-xk)$. We need $\gcd(b’,c_0+yk) = \gcd(a’,d_0-xk)=1$. Since $\gcd(a’,b’)=1$ from $S_a^- \cap S_b^- = \emptyset$ and $\gcd(y,b’)=\gcd(a’,x)=1$, we can generate $k_p$ for each $p\mid b’$ or $p\mid a’$ such that $p\nmid \gcd(b’,c_0+yk), p\nmid \gcd(a’,d_0-xk)$. Thus $k$ can be generated from $k_p$ and CRT and we thereby generate infinitely many $(m,n)$

CANBANKAN wrote: The answer is $P(x)=ax+b$ where there doesn’t exist a prime $p$ such that $\nu_p(a), \nu_p(b)$ are both positive, or both negative. Oops you missed that $(m,n)$ are positive integers, not integers. Then for example $P(x) = x+100$ would definitely not be a solution.

A fancy fix would be to say that $p=\infty$ is also a prime corresponding to the classical valuation of a number over the reals, thus $a$ and $b$ should also not be both positive, or both negative.

Tintarn wrote: A fancy fix would be to say that $p=\infty$ is also a prime corresponding to the classical valuation of a number over the reals, thus $a$ and $b$ should also not be both positive, or both negative. I was thinking about that too, but isn't the valuation at the place $\infty$ the log of absolute value (so it measures whether the number is in the interval $[-1,1]$)?