10. (10th US Olympiad Question) In a certain county, every two districts are connected by exactly one of three modes of transportation: car, train, or airplane. It is known that all three modes of transportation are used in the county, but no district has all three modes. Furthermore, no three districts are pairwise connected by the same mode of transportation. What is the maximum number of districts in this county?

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10. There are at most 4 districts.

Let vertices represent districts, and edges represent means of transportation, colored red, blue, and yellow to represent three types of transportation, as shown in the figure. This meets the requirements of the problem, thus there exist four districts.

If there are more than 4 districts, i.e., the graph has 5 vertices, divide them into non-red, non-blue, and non-yellow combinations. In this case, at least one combination will have $\geqslant 2$ vertices, otherwise, the number of vertices would be at most 3. We will discuss this in cases:
(i) Suppose the non-red combination has more than 3 vertices, then it contains three vertices $A, B, C$, which emit blue or yellow edges. Assume the non-blue combination has one vertex $D$. Then $D A$ is neither red nor blue, so it must be yellow. Similarly, $D B, D C$ are also yellow. In this case, if $A B$ or $A C$ is yellow, a monochromatic triangle is formed, which is a contradiction. Therefore, one of them must be blue, say $A B$ is blue. Similarly, $A C, B C$ are also blue, which results in $\triangle A B C$ being a blue triangle. This is a contradiction. Therefore, case (i) does not exist.
(ii) If each combination has at most 2 vertices, then since there are 5 vertices, one combination must contain two vertices. Suppose $A, B$ are in the non-red combination, and assume $A B$ is blue. Thus, there must be a point in the non-blue combination (otherwise, it would result in only two colors of edges), let this point be $C$. In this case, $A C, B C$ can only be yellow, and there must be another point, say $D$, in the non-yellow combination, which leads to $A D, B D$ being blue. Since $A B$ is blue, $\triangle A B D$ forms a monochromatic triangle, which is a contradiction. Therefore, case (ii) also does not occur.
Thus, a graph that meets the requirements of the problem cannot have 5 vertices.