We know that the graphs $K_5$ and $K_{3,3}$ are not planar.
The complete graph $K_5$ means that all 5 vertices are adjacent, thus requiring 5 different colors to allow coloring without two adjacent vertices having the same color.
The same applies to $K_{3,3}$ (a bipartite 3 by 3 graph), where each vertex is connected to not more than 3 other vertices.
Both $K_5$ and $K_{3,3}$ can be made planar by just taking away one single edge in each. Thus, 4 colors suffice to color the modified $K_5$, while the K3,3 is even 2-colorable.
Thus, it looks like the Four-Color Theorem would be proved by Kuratowski's theorem – no planar graph contains a $K_5$ nor a $K_{3,3}$.
Then, the biggest group of fully connected vertices would be 4 as in the modified $K_5$.
Could anyone explain why the above-mentioned condition is only necessary but not sufficient ?