I've seen this problem before in Donald J. Newman's "A Problem Seminar". Besides, no "assumption on the minimal value of $n$" can make the problem as it was posted true, since the points can be as close to each other as you want.

Let for a broken line $ M$ - $ M_{1},M_{2}, \ldots, M_{k}$, $ f(M)$ be the sum sum of the consequtive squares of its segments. It can be shown by induction on $ n$ that there exists broken line $ A,P_{1}, \ldots, P_{n}, B$ such that $ f(A,P_{1}, \ldots, P_{n}, B)\leq AB^{2}$. If $ n=0$ or $ n=1$ it is obvious! Suppose we proved this statement for $ n=0,1,...,k$. Lets examine the case $ n=k+1$. Draw the altitude $ CH$. If there are points in either triangle $ AHC$ and $ BHC$ then using the induction hypothesis $ f(A,P_{1}, \ldots, P_{t}, C)\leq AC^{2}$ and $ f(C,P_{t+1}, \ldots, P_{k+1}, B)\leq CB^{2}$. Using that $ P_{t}C^{2}+CP_{t+1}^{2}\geq P_{t}P_{t+1}^{2}$ (because $ \angle{P_{t}CP_{t+1}}$ is non-obtuse) and $ AC^{2}+BC^{2}=AB^{2}$ we prove the step $ n=k+1$ !!!! If all the points are inside one of triangles $ AHC$ and $ BHC$ then do the same for it. But this may happen only finite number of times because the triangle-s area tends to 0 while the c.o. of the points $ P_{1}, \ldots, P_{n}$ is constant. $ READY!$

you are wrong.

Why am I wrong? I don't understand. Can you explain what you mean please?

ttd wrote: you are wrong. Bump up. Why is his solution wrong, except that the argument with the triangles' areas tending to 0 needs a little clarification (in fact, it is also important that these triangles are all similar, so their diameters also tend to 0) ? Darij

ttd wrote: you are wrong. Why is it wrong @ttd ? I think it is a right solution.

Who can solve?