Clearly all constant $P$ work, and all linear $P$ with rational constant term work. It's also easy to see any other linear $P$ fail. Now suppose $\deg P \ge 2$; we will derive a contradiction. There must be some nonzero $r$ for which $P(r)$ is rational. By considering $Q(x) = P(rx)$, we can assume $r=1$. This implies $P(n)$ is rational for all positive integers $n$, so it follows $P$ has rational coefficients (eg. by Lagrange interpolation). Consider any nonzero $r'$ with $P(r')$ rational. It's easy to see $P(mr'+n)$ is rational for all positive integers $m,n$. Interpreting as a polynomial $Q$ in $m,n$, we deduce that $Q$'s coefficients are all rational, so if $P$ has leading term $a_dx^d$ then $a_d r'^i$ is rational for $0\le i\le d$; in particular it follows $r'$ is also rational. Pick some prime $q$ not dividing the denominator of any coefficient of $P$. Then there is some integer $q'$ not divisible by $q$ and some rational $s$ with $P(s) = \frac{q'}{q}$. It follows $s$ is of the form $\frac{t_1}{t_2q}$ for some integers $t_1,t_2$ with $(t_1,q)=1$. But then the denominator of $P(s)$ must be divisible by $q^2$, contradiction.

Call a polynomial contained if it satisfies the problem condition. First note that all constant polynomials are contained. Consider polynomials $P(x)=mx+c$ of degree $1$. We require that if $a=mx+c$ and $b=my+c$ are rational then so is $a+b-c$. This forces $c \in \mathbb{Q}$ and it's clear this is sufficient. We now show that no other polynomials $P$ are contained. Let $d=\deg{P} \geq 2$. Since $P$ is non-constant, we can choose some $\theta \in \mathbb{R}$ so that $P(\theta) \in \mathbb{Q}$. By the repeated application of the condition in the problem, $P{\left(n\theta\right)} \in \mathbb{Q}$ for each $n \in \mathbb{N}$. Let $Q(x)=P{\left(x\theta\right)}$ so $Q(n) \in \mathbb{Q}$ for each $n \in \mathbb{N}$. Performing Lagrange interpolation on $Q(0),Q(1),\ldots,Q(d)$ we see that $Q$ is a rational polynomial. Also note that for $x,y \in \mathbb{R}$: $$ Q(x),Q(y) \in \mathbb{Q} \implies P{\left(x\theta\right)},P{\left(y\theta\right)} \in \mathbb{Q} \implies P{\left((x+y)\theta\right)} \in \mathbb{Q} \implies Q(x+y) \in \mathbb{Q} $$so $Q$ is also contained. Next we show there is an irrational number $\alpha$ with $Q(\alpha) \in \mathbb{Q}$. Let $\frac{p}{q} \in \mathbb{Q}$ written in its simplest form. Write: $$ Q{\left(\frac{p}{q}\right)}=\frac{Q^{\prime}{\left(p,q\right)}}{Nq^{d}} $$where $Q^{\prime}$ is a homogeneous bi-variate integer polynomial of degree $d$ and $N$ is a positive integer. Let $c$ be the coefficient of $p^d$ in $Q^{\prime}$. We show that for any prime $r>\max{\left\{N,|c|\right\}}$, it $Q{\left(\frac{p}{q}\right)} \neq n+\frac{1}{r}$ for $n \in \mathbb{Z}$. If it were possible, this would mean: $$ N q^d (nr+1)=r Q^{\prime}{\left(p,q\right)} \implies r \vert q \xRightarrow{d \geq 2} r^2 \vert N q^d \implies r \vert Q^{\prime}{\left(p,q\right)} \xRightarrow{r \vert q} r \vert cp^d \xRightarrow{r>|c|} r \vert p $$which contradicts the fraction being in its simplest form. For $|n|$ sufficiently large, at least one of $n+\frac{1}{r}$, $-n+\frac{1}{r}$ will be in the image of $P$ and, as no rational inputs give this output, there must be $\alpha$ such that $Q(\alpha)$ takes one of these rational values. As $Q$ has rational coefficients, $Q(1) \in \mathbb{Q}$ and, by repeated application of the problem condition, $Q(\alpha+n) \in \mathbb{Q}$ for all $n \in \mathbb{N}$. Let $R(x)=Q(\alpha+x)$ then $R(n) \in \mathbb{Q}$ for $n \in \mathbb{N}$. By Lagrange interpolation on $R(0),R(1),\ldots,R(d)$ this means $R$ has rational coefficients. If $Q(x)=q_d \cdot x^d+q_{d-1} x^{d-1}+\cdots$ with $q_i \in \mathbb{Q}$ then: $$ R(x)=Q(\alpha+x)=q_{d} x^d+\left(d \alpha q_d+q_{d-1}\right) x^{d-1}+\cdots $$So $d \alpha q_d+q_{d-1} \in \mathbb{Q}$ which forces, as $q_d \neq 0$, $\alpha \in \mathbb{Q}$ which is a contradiction.

Sorry,I have a question.If p(x)=\sqrt(2)*x^2+3,then we can not find a rational number x(x is not equals 0),such that p(x) is rational number

@above: They first calibrate the original polynomial, so if $P(\theta)$ is rational, we consider $Q(x):=P(\theta x)$. Now, for the new polynomial you have $Q(1)$ is rational, and moreover $Q$ satisfies the same condition as $P$ - check it. This problem is quite similar to this one, thou the idea is a bit different.

Nice and cute! The answer is linear and constant $P$, which clearly work. Now we show that these are the only solutions. Assume $deg(P) \geq 2$, by the transformation $x \rightarrow rx$ WLOG $P(\mathbb{Z}) \in \mathbb{Q}$ so $P(\mathbb{Q}) \in \mathbb{Q}$, from this (say by lagrange interpolation) we know $P \in \mathbb{Q}[X]$ Claim 1: If $r \in \mathbb{R}$ s.t. $P(r) \in \mathbb{Q}$ then $r \in \mathbb{Q}$

Proof

Claim 2: For any polynomial of degree greater than $1$, there exists a non rational real number $c$ such that $P(c)$ is rational.

Proof

Hence we are done.