On reciprocal-sums of integer polynomials Let's consider the class $P \subset \mathbb{Z}[x]$ of nonconstant polynomials whose degree is at least two and with no roots in the positive integers. 
For $p \in P$, I'm interested in considering sums of the form $$S_{p,N} = \sum_{j=1}^{N} \frac{1}{p(j)}$$
$$\textbf{Question 1: Size}$$
If we assume the coefficients are nonnegative (with the leading coefficient positive), it's easy to deduce an absolute upper bound of $2$, by comparing with $\sum n^{-2} = \frac{\pi^2}{6}$. However, we allow negative coefficients, in which case it's not as clear if there is such a simple upper bound (at least not to me). 

What is $\sup\{S_{p,N}: (p,N) \in P \times \mathbb{N}\}$? Further, for a fixed degree $d \geq 2$, what is $\sup\{S_{p,N}: (p,N) \in P \times \mathbb{N} \ \text{and} \ \text{deg}(p) = d\}$?

I've tried Lagrange interpolation style arguments to prove that the former is $\infty$, but have not quite been able to make it work. 
$$\textbf{Question 2: Integrality}$$
For a given $p$, at most how many $N$ exist such that $S_{p,N}$ is an integer? Since $\sum_{j=1}^{\infty} \frac{1}{p(j)}$ converges and $p$ nonconstant, it's easy to see that there can only be finitely many such $N$.  
 A: Partial answer for 1: take the polynomial $p(x)=(x-1)^2(x-2)^2..(x-m)^2+1$. Then $p(k)=1, k=1,..m, p>0$ on the reals so no integer roots, no negative terms and obviously $S_{p,N} >  m$ for $N \ge m$, so definitely the supremum on all polynomials and all $N$ is infinity, while in degree $2m$ we see that the supremum is bigger than $m$ 
Edit later - actually we can even take $q(x)=(x-1)(x-2)..(x-m)+1$ since it is at least $1$ on the natural numbers and then we get that in degree $m$ the supremum is greater than $m$
A: Here is a proof that the supremum in fixed degree is finite, though my upper bound doesn’t feel particularly tight.  I found it simpler to consider the wider class of all integer polynomials of degree $\ge 2$ and the following variant sum:
$$S’_p := \sum_{\substack j\in \mathbb Z\\p(j)\ne 0} \frac1{|p(j)|}.$$
It’s easy to see by triangle inequality that $S’_p > |S_{p,N}|$, so it’s enough to show that $S’_p$ is bounded by a function of $\deg(p)$.  The extra restriction on $j$ doesn’t affect us on the domain $P$, since it can only exclude $j \le 0$.
We now consider the linear factorization of $p$ over $\mathbb C$:
$$p(x) = m(x-c_1)\cdots(x-c_d),$$
where $m$ is a non-zero integer, hence $|m| \ge 1$.  Let $a_i$ be the real part of $c_i$.  Then it is easy to see that for any real value of $x$,
$$|p(x)| \ge |m| \prod_{1\le i\le d} |x - a_i|.$$
Here is a simple but crucial observation:

For any integer radius $r\ge 1$, there are at most $2dr$ distinct integer values of $j$ which satisfy $|j-a_i| < r$ for some $i$.

For any $j$ outside of this set, we immediately have $|p(j)| \ge r^d$.  Let’s partition the index set into:
$$J_1 := \{ j \in \mathbb Z : \min_i |j - a_i| < 2\}, \\ J_2 :=  \{ j \in \mathbb Z : 2 \le \min_i |j - a_i| < 3\}, 
\\ J_3 :=  \{ j \in \mathbb Z : 3 \le \min_i |j - a_i| < 4\}, $$
and so on.
For any $j \in J_k$ with $k \ge 2$, we have $|p(j)| \ge k^d$.  For $j\in J_1$ we can use the trivial bound $|p(j)| \ge 1$ since $p(j)$ is an integer (recall that if $p(j)=0$ it’s excluded from $S’_p$), so the bound $k^d$ is uniform.  So we get the following upper bound:
$$S’_p \le \sum_{k=1}^\infty |J_k| k^{-d}.$$
The prior observation gives us upper bounds for the partial sums of $|J_k|$.  The fact that $k^{-d}$ is decreasing in $k$, combined with partial summation (or the rearrangement inequality) shows that the RHS achieves its maximum when we first maximize $J_1$, then $J_2$ with respect to our choice of $J_1$, and so on.
This gives the case $|J_1| = 4d$ and $|J_k| = 2d$ for all other $k$, so we finally have
$$S’_p \le 2d + 2d \sum_{k\ge 1} k^{-d} = 2d(1+\zeta(d)).$$
Since $\zeta(d)-1 \sim 2^{-d}$ for large $d$, this is essentially $4d + o(1)$.  As a concrete bound we can probably take $4d + 6/d$ for all $d\ge 2$.
