Assume all the people have distinct numbers of friends. Since the number of friends of someone ranges from $0$ to $n-1$, for each $k\in\overline{0,n-1}$ there is someone with $k$ friends. This means that the person with $n-1$ friends is a friend of everyone, including the person with $0$ friends, and this is a contradiction.

This is one of the first results which has to appear in every graph theory course : In any simple non-oriented graph, there are two vertices which have the same degree. Hmmm...maybe the second one. The first being the fact that the sum of the degrees is twice the number of edges.... Pierre.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally.

orl wrote: Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. I think grobber posted the solution yet, and there's no way to write it longer than that

Solved with popop614, hellohannah, qwerty123456asdfgzxcvb, OronSH, GrantStar, lrjr24, vsamc, peace09, MathJams. Otherwise, peoples' numbers of friends must be $0,1,\dots,n-1$ but it is impossible to have both $0$ and $n-1.$