Counting From the Complements

Let $N$ be the size of the group. Count pairs (person, colour) in which the colour is not that person’s favourite, in two ways.

By colour, the number of such pairs is the sum of the four tallies:

By person, each person has exactly one favourite among the four colours, so each person lies in exactly three of the four ‘not’ tallies. The number of pairs is therefore $3N$.

Equating the two counts gives $3N=75$, so $N=25$. As a check, the numbers preferring red, blue, yellow and green are $25-15=10$, $25-23=2$, $25-20=5$ and $25-17=8$. These are non-negative and sum to $25$, so such a group exists.

Key idea. With $k$ mutually exclusive and exhaustive categories, the complement tallies sum to $(k-1)N$.