Dividing 100 members into two groups. The question is:

An organisation has 100 members where each member is friends with exactly 56 other members (friendship is mutual). It is also known that there are 50 members who are friends with one another. Show that the 100 members can be divided into two groups such that for each group, each two members in it are friends.

I'll admit I haven't the slightest idea how to solve this question. My first thought was to use graph theory, but I don't think it would help to solve this question. Can anyone give me a clue?

Since there is a group of $50$ members who are friends of each other. Let's start with this group first.
Call this group "A" and its members $A_1, A_2, \cdots, A_{50}$.
Let the remaining $50$ members form a group $B$ with members  $B_1, B_2, \cdots, B_{50}$.
In group $A$, each member knows $49$ other $A$ group members. Because each $A_k$ knows exactly $56$ other group members, therefore each $A_k$ must know exactly $7$ $B$ group members.
For example $A_1$ may know $B_1, B_2, \cdots, B_7$ and no more.
Let's use the order pair $\left(A_i, B_k \right)$ to denote that $A_i$ and $B_k$ are friends, then there are $50 \times 7 = 350$ such order pairs.
Now consider the $B$ group. Each $B$ group member can know at most $49$ other $B$ group members. Thus each $B$ group members must know at least $7$ $A$ group members.
Thus from the perspective of $B$ group members, the number of order pairs $\left(A_i, B_k \right)$ is at least $50 \times 7 = 350$. Also if any $B$ knows more than $7$ $A$'s, the number of order pairs is greater than $350$.
Since the number of order pairs is exactly 350, each $B$ must know exactly 7 $A$'s and hence know exactly exactly $49$ other $B$'s.
In other words, any two $B$ group members are friends of each other.
