I claim the only answer is $f(x)=x+2010$. Note that, $x\in\mathbb{Q}^c\implies f(x)\in\mathbb{Q}^c$, that is, $x\in\mathbb{Q}\iff f(x)\in\mathbb{Q}$ (as $f(x)\in\mathbb{Z}[x]$). Now, let $f(x)=\sum_{k=0}^n a_k x^k$ with $a_n = 1$ and $a_0=2010$. Take a large prime $p>\sum_{k=0}^n |a_k|$. Now note that, $f(x)\to\infty$ as $x\to\infty$. In particular, $f(x)=p+2010$ has a solution, which happens to be a rational. By the rational root theorem, the only such solutions are $x\in\{\pm 1,\pm p\}$. From the choice of $p$, $f(x)=\pm 1$ is impossible. Also, by taking $p$ large, $-p$ case is also not possible. Hence, $x=p$, that is, $f(p)=p+2010$. Now since $p$ is arbitrary, we have that $f(p)=p+2010$ holds infinitely often, that is, $f(x)-(x+2010)$ has infinitely many distinct roots, so it must be a zero polynomial. This yields the answer above. Punchline: Any $f(x)\in\mathbb{Q}[x]$ with $f^{-1}(\mathbb{Q})\subseteq \mathbb{Q}$ is necessarily linear. This is originally a problem from an old Iranian olympiad.

grupyorum wrote: I claim the only answer is $f(x)=x+2010$. Note that, $x\in\mathbb{Q}^c\implies f(x)\in\mathbb{Q}^c$, that is, $x\in\mathbb{Q}\iff f(x)\in\mathbb{Q}$ (as $f(x)\in\mathbb{Z}[x]$). Now, let $f(x)=\sum_{k=0}^n a_k x^k$ with $a_n = 1$ and $a_0=2010$. Take a large prime $p>\sum_{k=0}^n |a_k|$. Now note that, $f(x)\to\infty$ as $x\to\infty$. In particular, $f(x)=p+2010$ has a solution, which happens to be a rational. By the rational root theorem, the only such solutions are $x\in\{\pm 1,\pm p\}$. From the choice of $p$, $f(x)=\pm 1$ is impossible. Also, by taking $p$ large, $-p$ case is also not possible. Hence, $x=p$, that is, $f(p)=p+2010$. Now since $p$ is arbitrary, we have that $f(p)=p+2010$ holds infinitely often, that is, $f(x)-(x+2010)$ has infinitely many distinct roots, so it must be a zero polynomial. This yields the answer above. Punchline: Any $f(x)\in\mathbb{Q}[x]$ with $f^{-1}(\mathbb{Q})\subseteq \mathbb{Q}$ is necessarily linear. This is originally a problem from an old Iranian olympiad. I think even $cx+2010$ works where $c$ is integer. @below oh sorry

Read the question carefully. Leading coefficient is one.