nt.number theory - A special integral polynomial

Yes, it is always possible.

First note that it suffices to construct a totally complex Galois extension $K/mathbb{Q}$ of degree $2n$ with Galois group $S_{2n}$. By the primitive element theorem, this extension is of the form $mathbb{Q}[t]/(f(t))$ for some irreducible polynomial $f$, the minimal polynomial of an algebraic number $alpha in K$. Then there exists $n in mathbb{Z}^+$ such that $n alpha$ is an algebraic integer: take the minimal polynomial of that algebraic integer: it generates the same field extension.

To construct the desired extension $K$, in turn it suffices to find an irreducible polynomial with $mathbb{Q}$-coefficients with no real roots and whose Galois group is the largest possible $S_{2n}$. This is possible by a weak approximation / Krasner's Lemma argument. I will just sketch it for now; I can fill in more details if needed. The idea is to find a finite set of primes $p$ and degree $2n$ polynomials $f_p$ such that the Galois group of $f_p$, as a group of permutations on the roots of $f_p$, is of a certain form (e.g. contains a specific transposition). Also let $f_{infty}$ be any degree $2n$ polynomial over $mathbb{R}$ without real roots. Then by Krasner's Lemma, there exists a polynomial $f$ which is sufficiently $p$-adically close to each $f_p$ and to $f_{infty}$ to have the same local behavior: in particular, to factor the same way over $mathbb{Q}_p$ and over $mathbb{R}$ and to generate the same local Galois groups. Then, by identifying the local Galois groups with decomposition groups at $p$ (of unramified extensions), if one has enough primes so as to get permutations of every possible cycle type, then the global Galois group of $f$ certainly must be $S_{2n}$. Indeed, to see this we use the following result from lecture notes of Keith Conrad (and Bertrand's postulate!):

http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/galoisSnAn.pdf

Theorem: For $n geq 2$, a transitive subgroup of $S_n$ which contains a transposition and a $p$-cycle for some prime $p > frac{n}{2}$ is $S_n$.

The condition at infinity means that $mathbb{Q}[t]/(f(t))$ is totally complex, hence so is its splitting field. To ensure that $f$ is irreducible, we may apply Krasner's Lemma again and take its coefficients sufficiently close to those of an irreducible degree $2n$ polynomial over $mathbb{Q}_p$ (for a different $p$ from those used thus far) so as to be irreducible over $mathbb{Q}_p$, which implies irreducibility over $mathbb{Q}$.

This can in principle be made explicit, but I might search the literature for a known classical family of polynomials doing what you want before I tried to carry out this construction explicitly.