Does there exist polynomial $p(x)$ with integer coefficients such that $p(x) > 0$ for all real values of $x$, but for any integer $n > 0$ there exists integer $k$ such that $n$ divides $p(k)$? The trick with multiplying quadratic polynomials (as here Diophantine equation with no integer solutions, but with solutions modulo every integer) does not work, if I am not mistaken.

$\newcommand{\Q}{\mathbf{Q}}\newcommand{\Z}{\mathbf{Z}}$ I now suspect the answer might be no! This isn't a complete answer but it might be an idea that turns into one. So let me assume that such $p$ exists and let me go for a counterexample. First I claim that if such $p$ exists, then a monic $p$ exists. For if $p$ works, then so does $Np$ for a large positive integer $N$, and if $p=cx^d+\ldots$ and $N=c^{d1}$ then $Np=q(cx)$ with $q$ monic with integer coefficients, and $q$ also works. So WLOG $p$ is monic. Now say $p$ factors as $p_1p_2\ldots p_r$ in $\Q[x]$ with the $p_i$ monic. By Gauss' Lemma the $p_i$ are all in $\mathbf{Z}[x]$. Note that $p$ must have positive degree so $r\geq 1$. Let $K$ be the splitting field of $p$ and let $K_i$ be the field $\Q[x]/(p_i)$. Now none of the $K_i$ have any real places. Let $c$ denote any complex conjugation in $Gal(K/\Q)$. By Cebotarev density, there's a prime number $\ell$, unramified in $K$, and such that $Frob_\ell$'s conjugacy class in $Gal(K/\Q)$ is that of $c$, and indeed there are infinitely many such $\ell$. Certain primes cause me trouble, so let me get rid of them now: for each $K_i$ let $S_i$ be the index of $\Z[x]/(p_i)$ in the full ring of integers of $K_i$, and let $S$ be the product. Now let $\ell$ be a prime as above, and such that $\ell$ doesn't divide $S$. I claim that this $\ell$ is going to cause us problems. Say $p(k)$ is zero mod $\ell$. Then one of the $p_i(k)$ is zero mod $\ell$, so the factorization of $p_i$ mod $\ell$ has a linear factor, and so $\ell$ factors in the integers of $K_i$ into a bunch of primes one of which has degree 1. I now want to argue something like the following: the Frobenius element at $\ell$ corresponding to this prime is fixing a root of $p_i$ so complex conjugation is fixing a root of $p_i$ so $p$ has a real root! But the cricket has just started and I can't think straight. Can this be turned into a proof? 

