Compute $\sum_{x\in\mathcal{N}}\sum_{y\in\mathcal{N}}\frac{1}{x+y+1}$ 
Let $n$ be a natural number with $3\nmid n$. Let $\mathcal{I}$ be the set of all $n$-th roots of unity. Then calculate $$\sum_{x\in\mathcal{I}}\sum_{y\in\mathcal{I}}\frac{1}{x+y+1}$$

I tried to simplify the form of the problem by expressing the roots through the primitive root (say $y_0$) as follows $$\sum_{j\in(\overline{0,n-1})}\sum_{k=0}^{n-1}\frac{1}{y_0^j+y_0^k+1}=\sum_{j\in(\overline{0,n-1})}\sum_{k=0}^{n-1}\frac{1}{y_0^j+y_0^k+y_0^n}$$ and was going to come down to further workable form, by factoring every time the term $y_0^{min(k,j,n)}$ in the denominator. But some tests vaguely show that this process may last periodically with no any useful result...
Any help is appreciated.
 A: Notice for any $z \in \mathcal{N}$, the map $\mathcal{N} \ni x \mapsto xz\in \mathcal{N}$ is a permutation over $\mathcal{N}$. We have
$$\sum_{x,y \in \mathcal{N}} \frac{1}{x + y + 1}
= \sum_{x,y \in \mathcal{N}} \frac{z}{zx + zy + z}
= \sum_{x,y \in \mathcal{N}} \frac{z}{x + y +z }
$$
Take average over $z$, we obtain
$$\sum_{x,y \in \mathcal{N}} \frac{1}{x + y + 1}
= \frac1n \sum_{x,y,z\in\mathcal{N}} \frac{z}{x+y+z}$$
By symmetry among $x,y,z$, this leads to
$$\sum_{x,y \in \mathcal{N}} \frac{1}{x + y + 1}
= \frac1{3n}\sum_{x,y,z\in\mathcal{N}}\frac{x + y + z}{x + y + z} = \frac1{3n} n^3 = \frac{n^2}{3}$$
By a similar argument, it is trivial to show
$$\mathcal{I}_k \stackrel{def}{=}\sum_{x_1,\ldots,x_k \in \mathcal{N}}\frac{1}{1+ \sum_{j=1}^k x_j}
= \sum_{x_1,\ldots,x_k \in \mathcal{N}}\frac{z}{z + \sum_{j=1}^k x_j}, \forall z \in \mathcal{N}
$$
As long as we are working with a $n$ where $\sum_{j=1}^k x_j = -1$ doesn't have a solution in $\mathcal{N}^k$, this leads to
$$\mathcal{I}_k
= \frac{1}{(k+1)n}\sum_{z,x_1,\ldots,x_k \in \mathcal{N}}\frac{z + \sum_{j=1}^k x_j}{z+ \sum_{j=1}^k x_j}
= \frac{1}{(k+1)n} n^{k+1} = \frac{n^k}{k+1}$$
