$P(x)= \frac{x(x+1)}{2} +1$. 

It is easy to see that $P^{n+1}(0) > P^n(0)$ and $P$ maps the integers into the integers.

But I think (didn't check it, migth be one of these facts which are obvious but wrong) that 

$$P^(n)(x) = \frac{1}{2^{m} x^{2^n}+....\notin \ZZ $$

where $m$ is probably $m=2^n+1$.

The right question to ask migth be if $f$ maps the integers into the integers....

EDIT: P.S. The answer with the rationals turns out to be true, I think (my algebra is rusty):

Let $P$ be such a polynomial, and let $m$ be the degree of $P$. Then using teh Lagrange interpolation formula, you can reconstruct $P(x)$ from $m+1$ distinct integer values of the type $P^{(k)}(0)$, and since all of these are rational, all the coefficients are rational. Actually this way one can prove the following Lemma:

**Lemma:** If $deg(P)=m$ and there exists $m+1$ distinct rational numbers $x_1,..., x_{m+1}$ so that $P(x_1),...,P(x_{m+1}$ are all rational, then $P(X) \in \QQ[x]$.

Lagrange formula also explains why in the first case we can only get rational integers.