III OBM

Suppose that no point is joined to all the others. Then given any point $X$ we can find $Y$ not joined to $X$. So take arbitrary $A$ and $C$. Then take $B$ not joined to $A$ and $D$ not joined to $C$. Then the four points $A,B,C,D$ do not meet the required condition. Contradiction.

So find $X_1$ joined to all the other 99 points. Now repeat the argument for the other 99 points, that gives a point $X_2$ joined to the other 98. But it is also joined to $X_1$, so it is joined to all other 99 points. Now repeat for the other 98 points and so on. The last time we can repeat is when we have already found $X_1,X_2,\dots ,X_{96}$ leaving four points. We can now take $X_{97}$ joined to the other three and hence to all other 99. Thus we can get at least 97 points each joined to all points except itself.