1990 APMO 第 4 题

A set of 1990 persons is divided into non-intersecting subsets in such a way that
(a) no one in a subset knows all the others in the subset;
(b) among any three persons in a subset, there are always at least two who do not know each other; and
(c) for any two persons in a subset who do not know each other, there is exactly one person in the same subset knowing both of them.
(i) Prove that within each subset, every person has the same number of acquaintances.
(ii) Determine the maximum possible number of subsets.

Note: it is understood that if a person $A$ knows person $B$, then person $B$ will know person $A$; an acquaintance is someone who is known. Every person is assumed to know one's self.

一组 1990 个人被分成不相交的子集，方式如下：

(a) 子集中没有人知道子集中的所有其他人；

(b) 在一个子集中的任意三个人中，总是至少有两个人彼此不认识；和

(c) 对于子集中任意两个彼此不认识的人，同一子集中恰好有一个人认识他们两个。

(i) 证明在每个子集中，每个人都有相同数量的熟人。

(ii) 确定子集的最大可能数量。

注意：可以理解，如果一个人$A$认识一个人$B$，那么这个人$B$就会认识一个人$A$；熟人是认识的人。每个人都被认为了解自己。