At the end, did the blackboard contain exactly the integers between $-2016$ and $2016$, or did it contain these integers and perhaps some other ones?

There could be some other numbers

$N = 2$ suffices. Let $A$ be a positive integer divisible by $1\sim 2016$ and suppose $0$ and $A$ are initially on the blackboard. First consider the polynomial $Ax + A$. This allows us to add the root $x = -1$ to the board. Then consider $-x^{A} - x^{A-1} - \dots - x + A$, which gives a root $x = 1$. Now suppose integers with absolute value less than $k$ are already on the board. Write $\frac{A}{k}$ in base-$k$ as $(b_n \dots b_1)_{k}$. Then the polynomial $- b_n x^n - b_{n-1} x^{n-1} - \dots - b_1 x + A$ yields a root $ x = k$. Similarly we can obtain $-k$ by changing the signs of the $b$s appropriately. Thus we can inductively obtain all integers between $-2016$ and $2016$. In fact, it is sufficient to have $A$ divisible by every prime number in $1 \sim 2016$. For that we instead work with $\frac{Ad}{k}$ where $d = \frac{k}{rad(k)} < k$. This ensures that $\frac{Ad}{k}$ is an integer even when $\frac{A}{k}$ is not. Suppose in base-$k$ we have $\frac{Ad}{k} = (c_n \dots c_1)_{k}$, then the polynomial $- \frac{c_n}{d} x^n - \frac{c_{n-1}}{d} x^{n-1} - \dots - \frac{c_1}{d} x + A$ has root $x = k$. Furthermore, we can have $-\frac{c_j}{d}$ on the board because it is the root of $dx + c_j$, and $d, c_j$ are both less than $k$. On the other hand, it is also necessary that $A$ be divisible by every prime less than $2016$. For that we use the following lemma: Lemma. Suppose $D$ is a positive integer and $a_n, a_{n-1}, \dots, a_0$ are complex numbers such that for each $j$, $Da_j$ is an algebraic integer and if $a_j \neq 0$ then $\frac{D}{a_j}$ is also an algebraic integer. Take any root $\lambda$ of the polynomial $a_n x^n + \dots + a_0$. Then $D^2 \lambda$ is an algebraic integer, and if $\lambda \neq 0$ then $\frac{D^2}{\lambda}$ is also an algebraic integer. Proof. We have $a_n(D^2 \lambda)^{n} + a_{n-1}D^2 (D^2\lambda)^{n-1} + \dots + a_0D^{2n} = 0$. Equivalently $(D^2 \lambda)^{n} + \frac{a_{n-1}D^2}{a_n} (D^2\lambda)^{n-1} + \dots + \frac{a_0D^{2n}}{a_n} = 0$. The coefficient $\frac{a_{n-k}D^{2n-2k}}{a_n}$ is the product of $a_{n-k}D$, $\frac{D}{a_n}$ and $D^{2n-2k-2}$, each of which is an algebraic integer. Therefore the number $D^2\lambda$ satisfies a monic equation all of whose coefficients are algebraic integers. This means $D^2 \lambda$ is an algebraic integer as desired. The argument for $\frac{D^2}{\lambda}$ is symmetric and thus omitted. With this lemma, it is not hard to see that if $A$ is not divisible by $p$, then we can never write $p$ on the board.