5.60 At a party, the way every two people greet each other is one of four ways (nod, handshake, kiss, hug). Kandi and Rendi kissed, but Kandi did not kiss Sadi. For every three people, the ways they greet each other are either all the same or all different. How many people can there be at this party at most?
[Solution] At most 9 people.
Represent each person with a point, and color the line connecting two points according to the way they greet each other in one of four colors: red, yellow, blue, or black.

Let Kondi be point $V_{1}$. If the number of people is not less than 10, then among the $q$ lines drawn from point $V_{1}$, there must be $\left[\frac{q}{4}\right]+1=3$ lines of the same color. Suppose $V_{1} V_{2}, V_{1} V_{3}, V_{1} V_{4}$ are red.

According to the given condition, the three sides of $\triangle V_{1} V_{2} V_{3}, \triangle V_{1} V_{3} V_{4}, \Delta V_{1} V_{2} V_{4}$ are all red, thus the three sides of $\triangle V_{2} V_{3} V_{4}$ are all red.

According to the known condition, there exists a point $V_{5}$ such that $V_{1} V_{5}$ is not red. Let $V_{1} V_{5}$ be yellow. Since the three sides of $\triangle V_{1} V_{2} V_{5}$ are all different colors, $V_{2} V_{5}$ can only be blue or black.
Similarly, $V_{3} V_{5}, V_{4} V_{5}$ can only be blue or black.
Thus, among $V_{2} V_{5}, V_{3} V_{5}, V_{4} V_{5}$, at least two sides are of the same color. Without loss of generality, assume $V_{2} V_{5}$ and $V_{3} V_{5}$ are both blue. In this case, for $\triangle V_{2} V_{3} V_{5}$, there are two sides that are blue and one side that is red, which contradicts the given condition.
Therefore, the number of people is no more than 9.
Furthermore, it is easy to verify that 9 people are possible.