NP-Completeness of a Graph Coloring Problem
It is quite simple to show that the alternative formulation is NP-hard. The reduction is from the vertex coloring problem. Given a graph G with $n$ vertices, we create an instance of the above problem with $n$ output vertices and $n$ input vertices. Weights are set as follows: For all $i$, let $w_{ii}=1$. For $i \neq j$, if there is an edge between vertex $i$ and vertex $j$, let $w_{ij}=w_{ji}=1$, else let $w_{ij}=w_{ji}=0$. In addition, let $\beta=1$.

This is quite obvious but difficult to describe why the reduction is correct.
Let $\mathcal{C}$ show the instance of the graph coloring and $\mathcal{R}$ show the reduced instance of the problem.
To show the above reduction gives a correct solution we need to show that 
(1) every valid coloring for $\mathcal{R}$ is valid for $\mathcal{C}$ as well.
(2) the answer given by $\mathcal{R}$ is minimal for $\mathcal{C}$.

If $i$ and $j$ are two adjacent vertices of $\mathcal{C}$,
then they must have different colors in $\mathcal{R}$.
That is because if $i$ and $j$ are adjacent and they have the same color,
the fraction $\frac{w_{jj}}{1+\sum_{c(i)=c(j),i \neq j} w_{ij}}$ would result in $\frac{1}{1+X}$, 
where $X$ has a positive value. 
Therefore, the condition does not hold.
In addition, every valid (but not necessarily minimal) coloring for $\mathcal{C}$,
is a valid coloring for $\mathcal{R}$ as well.
It follows the fact that in a valid coloring of $\mathcal{C}$, 
every pair of adjacent nodes have different colors, 
so the condition holds for all the colored edges of $\mathcal{R}$ in the solution.
Since every coloring of $\mathcal{C}$ is a valid coloring for $\mathcal{R}$,
the minimal solution to $\mathcal{R}$ must be a minimal solution to $\mathcal{C}$
as well. 
Otherwise, it is not a minimal one because the solution to $\mathcal{C}$
gives a solution with a smaller number of colors.