Given that $G_n = (V, E)$ is the graph obtained by deleting one edge from the complete graph $K_n$, we aim to find the chromatic number $\chi(G_n)$, which is the smallest number of colors required to color the vertices such that no two adjacent vertices share the same color.

Analysis:

1. Complete Graph $K_n$: The chromatic number of a complete graph $K_n$ is $\chi(K_n) = n$ because every vertex is connected to every other vertex, meaning each vertex must have a unique color.

2. Graph $G_n$: By deleting one edge from $K_n$, the graph $G_n$ will still have a high level of connectivity. Specifically:

   • For $n = 3$, removing an edge from $K_3$ gives a path (or a 2-edge graph), which can still be colored with 2 colors. Hence, $\chi(G_3) = 2$.
   • For $n \geq 4$, even though we remove one edge, the remaining graph is still highly connected. The graph does not become bipartite, and all vertices will still need distinct colors due to their near-complete connectivity. Thus, $\chi(G_n) = n$.

Conclusion:

• For $n = 3$, $\chi(G_3) = 2$.
• For $n \geq 4$, $\chi(G_n) = n$.

Would you like more details on the reasoning? Also, here are some related questions:

1. What is the chromatic number of a graph with one edge removed from a bipartite graph?
2. How does edge deletion affect the chromatic number in general?
3. How does the chromatic number relate to other graph invariants like the clique number?
4. What are examples of graphs where edge deletion significantly reduces the chromatic number?
5. Can the chromatic number be less than the clique number in some graphs?

Tip: When working with chromatic numbers, it's often useful to consider how much structure (connectivity) is retained after edge deletions.