A set of integer roots of polynomial Inspired by the problem 9 of a team contest (where $2016$ was replaced by $2014$), I would like to know what can be said about the set of integers (positive and negative ones) $S$ knowing the following two rules:


*

*The numbers $0$ and $2016$ are in $S$.

*If an integer $n$ is a root of a non-zero polynomial with coefficients in $S$, then $n\in S$.


Can we describe all the numbers that are in $S$ for this $2016$-case? (and for the $2014$-case?) 
 A: Let's agree upon the additional constraint suggested by Element118 in the comments.
Let $N=2016$. Most of what follows works for all natrurals $N$, some for even $N$, some only for more specific $N$. The most specific result ("The minimal $S$ contains precisely $0$ amnd the divisors of $N$") follows at least for $N=2016$ and also for some $N$ with very strict (but not explicitly formulated) constraints for its prime factorization.
We have $$\tag1-1\in S$$ as it is a root of $N x + N$. After that we have $$\tag21\in S$$ from $-x^{N}-x^{N-1}-\ldots -x+N $ (works for every "year $N$")
and  $$\tag32\in S$$ from $-x^{10}-x^9-x^8-x^7-x^6-x^5+2016$ (this is obtained from the binary representation of $2016$ and works similarly for any even $N$).
Also, 
$$\tag4 n\in S\implies -n\in S.$$
(This follows from the polynomial $x+n$)
and
$$\tag5n,m\in S, n\mid m\implies \frac mn\in S$$
(from $-m\in S$ and the polynomial $nx+(-m)$).
Equipped with $-1$ and $-2$ we can try the same trick with base $3$ (at least for $2016$, which is a multiple of $3$): We find $$\tag63\in S$$ from $-2x^6-2x^5-2x^3-2x^2+2016$ (because $2016=2202200_3$; any year a  multiple of $6$ would work).
As $4\mid 2016$ we next use $2016=133200_4$ to find $4\in S$ from $- x^5-3x^4- 3x^3-2x^2+2016$. If we had $5\mid 2016$ we could continue like this, but, alas, we can't. 
Indeed, we have
Lemma. For the minimal $S$ with the given properties, we have that $0\ne n\in S$ implies $n\mid 2016$.
Proof.
Let $S_0=\{0,N\}$ and then recursively $S_{k}$ the union of $S_{k-1}$ with all integer roots of non-trivial polynomials with coefficients in $S_{k-1}$. Then $S:=\bigcup_k S_k$ is clearly the minimal set with the desired property: It does have the desired property and everything we added to it needed to be added.
We show by induction on $k$ that $0\ne n\in S_k$ implies $n\mid N$, with the case $k=0$ being trivial. So let $k>0$ and $n\in S_k\setminus S_{k-1}$. By definition of $S_k$ there exists a non-zero polynomial $p(X)=a_0+a_1X+\ldots+a_dX^d$ with coefficients in $S_{k-1}$ and with $p(n)=0$. Because $n\ne 0$ we may assume wlog that $a_0\ne 0$. But then $a_0=-a_1n-\ldots -a_dn^d$ shows that $a_0$ is a multiple of $n$. As $a_0$ is also a divisor of $N$ by induction hypothesis, so is $n$.
This completes the induction proof, whence the claim follows. $\square$
Specifically, for the year 2016 we can only expect numbers of the form $\pm2^a3^b7^c$ with $0\le a\le 5$, $0\le b\le 2$, $0\le c\le 1$ to occur in $S$.
From what we already have ($2,3,2016\in S$) and $(5)$ we find that
$\pm 2^a3^b7\in S$ for $0\le a\le 5$, $0\le b\le 2$. Specifically $7\in S$ so that also $\pm 2^a3^b\in S$ and thus all divisors of $2016$ are in $S$. 
By the lemma, this is as much as we can expect.

Note that we were somewhat lucky to find two of three prime divisors of $2016$ "by hand"; I don't think the same works for $2014$, not to mention $N$ with more prime divisors. I suppose that at least $N$ of the form $N=q\cdot\prod_{i=1}^kp_i^{a_i}$ works if all numbers $\le \max\{p_i\}$ are divisors of $N$: Using the base expansion trick as exercised above, we find that all $p_i$ are $\in S$, then by $(5)$ also $q\in S$ and thus by $(5)$ again all divisors.
A: Considering the polynomial
$$2016x+2016$$
we get $-1 \in S$. This allow us to consider the polynomial
$$2016x^{2016}-x^{2015}-...-1$$
and we get $1 \in S$. Now just write $2016$ in base $2$ : $11111100000$, and take the corresponding polynomial
$$x^{10}+x^{9}+x^8+x^7+x^6+x^5-2016$$
and you get $2 \in S$. Now that you can use the coefficient $2$, you can write $2016$ in base $3$ and do the same thing, so $3 \in S$ and for the same reason $4 \in S$. Since $2016$ is not divisible by $5$, you can't do the same because you have a remainder. Notice also that if $a \in S$, then $-a \in S$, since you can take $x+a$. 
You can do the same thing with $2014$, but you can't get $3$ since $2014$ is not divisible by $3$. I'm not sure though if there is another method to get other numbers in $S$.
