1. Graph Representation:
- Represent the chess tournament as a graph $G$ where each vertex represents a contestant.
- Draw an edge between two vertices if the corresponding contestants did not play each other.

2. Properties of the Graph:
- Given that if any two contestants $A$ and $B$ did not play each other, there exist exactly two other contestants $C$ and $D$ who played against both $A$ and $B$. This implies that for every edge $AB$ in $G$, there is a pair of vertices $C$ and $D$ such that $AC, AD, BC,$ and $BD$ are not edges in $G$. We call such pairs of edges $\textit{conjugates}.$
- Additionally, no four contestants played exactly five games among them, which implies that there are no subgraphs isomorphic to $K_4$ minus one edge.

3. Maximal Degree Argument:
- Consider a vertex $X$ with maximal degree in $G$.
- Let $U$ be a neighbor of $X$ and $Z$ be an endpoint of the conjugate edge of $XU$.
- For each neighbor $Y$ of $X$, either $ZY$ is an edge, or $Z$ is the endpoint of the conjugate edge of $XY$. This implies that $Z$ has at least as many neighbors as $X$, so $Z$ also has maximal degree.

4. Breadth-First Search (BFS) Argument:
- Using BFS, we can show that every vertex in the graph has the same degree.
- Start from a vertex with maximal degree and explore its neighbors. By the previous argument, all these neighbors must also have maximal degree.
- Continue this process iteratively. Since the graph is connected, every vertex will be reached, and each will be shown to have the same degree.

5. Conclusion:
- Since every vertex in the graph has the same degree, it follows that every contestant played the same number of games.

$\blacksquare$