Let $A,B$ be two people who do not know each other. They have exactly one common acquaintance $C$, and for each acquaintance $D_a\ne C$ of $A$ there is exactly one common acquaintance $D_b\ne C$ of $D_a,B$. This way, we can pair up the acquaintances of $A$ and those of $B$ which are different from $C$, and the conclusion is that every two people who do not know each other have the same number of acquaintances. In the graph $G$ which we associate to this situation in an obvious manner, two vertices can have different degrees only if they lie in different connected components of the complementary graph $\overline G$. Now, if $\overline G$ has at least two components each of which has $\ge 2$ vertices, there will be $4$ vertices $a_1,a_2,b_1,b_2$ such that $a_i,b_j$ are connected for $i,j=1,2$, which contradicts the hypotheses, and if it has at least two components, one of which has exactly one vertex, then that vertex is connected to all the others, so we get a contradiction again. This means that $\overline G$ is connected, and it follows from this and the first sentence in the paragraph that all vertices of $G$ have the same degree. It's easy to see that there are no graphs $G$ with the required properties and with fewer that $5$ vertices, and that a $5$-cycle satisfies all the conditions; it follows from this that the largest possible number of subsets is achieved when we divide the set $1990$ vertices (well, people ) into $398=\frac{1990}5$ subsets.

grobber wrote: Now, if $ \overline G$ has at least two components each of which has $ \ge 2$ vertices, there will be $ 4$ vertices $ a_1,a_2,b_1,b_2$ such that $ a_i,b_j$ are connected for $ i,j = 1,2$, which contradicts the hypotheses, Sorry to revive this old thread. I don't understand this part of the solution. Can you explain why is it a contradiction?

A square in this graph either requires an additional edge (so a triangle) or has two disconnected vertices with two common neighbors (contradicting the "exactly 1" in condition 3).

I'm fairly sure such a graph must be a 5 cycle or 2 attached 5 cycles. if any vertex has degree less than 2, since it can't be zero it has degree one but then the one it is attached to is itself attached to all others since that first one is not attached to any =><= Now since each vertex has degree 2 we have a cycle. We clearly do not have a 4 or 3 cycle but if the length is 6 or greater take 2 verticies 2 apart if they are connected we have a smaller cycle if not then they are connected to another which if put together with the part of the cycle that vertex is not part of gives us a smaller cycle so we have a 5 cycle. Every vertex not connected to the 5 cycle that we will call ABCDE is connected to at most of them. We consider pairs X, Y with X not in ABCDE and Y in ABCDE. And note that X,Y have a common neighbor outside ABCDE if and only if they are not connected X is not connected to a neighbor of Y in ABCDE. If any vertex is not attached to any of ABCDE it contributes 5 pairs while those attached to one contribute 2 since it is not possible for a vertex to be common neighbors for three such pairs ( if two are connected to one vertex then those 2 have two common neighbors one outside ABCDE and one of ABCDE they are both connected to) thus we only consider the case where each vertex outside is attached to exactly one of ABCDE and each is a common neighbor of exactly 2 pairs X,Y. So each vertex outside ABCDE has degree 3. Now if a vertex is connected to A then we need another vertex to connect it to C going around like this we find there is a vertex connected to each of ABCDE. Now consider a neighbor of A and one of C if they are not connected then their neighbor T is needs to be connected to a vertex of ABCDE which is not a neighbor of A or C which is impossible thus each neighbor of A is connected to each neighbor of C. Thus there is at most 1 neighbor of C so C has degree 2+1=3 so we are done. If the 5 cycle is the entire graph we are also done.