Special Vertex Partitioning When can we partition the vertices of a graph $G$ into $n$ subsets such that every vertex is adjacent to vertex from every subset? For example, in the following graph, we have partitioned the vertices into $2$ subsets.

Is anything known about this type of partitioning? A motivation for studying these partitionings is that we can represent vertices as possible states and adjacent vertices as states reachable through one move. Then this partitioning allows us to convey certain messages no matter what the state.
 A: I realized that that in my old answer I misunderstood the question. 
Let $N’(G)$ be the maximum $n$ for which such a partitioning exists. Then it is clear that such a partitioning exists also for all $n\le N’(G)$. Now I can only say that $N’(G)\le\delta(G)$, where $\delta(G)$  is the minimum vertex degree of the graph $G$. 

An old version.
Let $N(G)$ be the maximum $n$ for which exists a partitioning of the set of vertices of the graph $G$ such that each two distinct parts contain adjacent vertices. Then it is clear that such a partitioning exists also for all $n\le N(G)$. We have the following simple bounds for $N(G)$.
If $|V(G)|\ge 1$ then $\chi(G)\le N(G)\le V(G)$,  where $V(G)$ is number of vertices of the graph $G$ and $\chi(G)$ is a chromatic number of the graph $G$, that is the minimal number of colors needed to color all vertices of the graph $G$ such that there are no adjacent monochromatic vertices. 
$m(G)\le  N(G)(N(G)-1)/2\le E(G)$, where $E(G)$ is number of edges of the graph $G$ and $m(G)$ is a number of edges of a maximum matching of the graph $G$, that is the maximal number of mutually non-adjacent edges of the graph $G$.
The upper bound $N(G)\le V(G)$ seems to be a corollary of the bound $N(G)(N(G)-1)/2\le E(G)$, because $E(G)\le V(G)(V(G)-1)/2$ (I assumed that the graph $G$ has no double edges).
The upper bounds are far from optimal, because a graph $G$ with big both $V(G)$ and $E(G)$ can have small $N(G)$. For instance, if $S_n$ is the star with $n$ vertices (the tree with one root and $n-1$ leaves) then $V(S_n)=n$, $E(S_n)=n-1$, whereas $N(S_n)=2$.
