This is essentially a version of IMO 2015 problem 1, meaning I read that problem wrong, and thought it was this problem. They intended $AB=AC$ to mean that the distances were the same, I thought it meant they were the same line (i.e. A,B,C are colinear). I eventually realized that this wasn't the same problem, but I still want to know if there is a proof. As I "came up with it" myself, there is no proof online yet, so I thought I'd ask here. I will often use "set" to mean "set of points in a plane" in this post. I am not asking for a solution to IMO 2015 problem one, but this bastardized version instead.
A more rigorous way to pose the question would be this: A "balanced" set $S$ is one where for any two elements $A$ and $B$ in $S$, there exists a $C$ in $S$ on the line $AB$. An "uncentered" set $S$ is one where for any two elements $A$ and $B$ in $S$, there are no $C$ and $D$ in $S$ both on the line $AB$. For which positive integers $n\ge3$ is there a set $S$ of size $n$ that is both balanced and uncentered?
Essentially, all lines through points in $S$ have $3$ points, never less, never more.
My conjecture is that the only possible value of $n$ is $3$ ($S$ consists of 3 colinear points), but I could be wrong.
This is some of my work:
Call $L$ the set of all lines through 2 or more points of $S$, and $L_p$ the subset of $L$ containing all lines through $p$
If $n>3$, all points being colinear no longer works, as there would be $4$ colinear points, which isn't allowed. This trivially means that $\lvert L\rvert\ge3$. I won't prove this here but it's very simple.
Now, lets try to find what $\lvert L\rvert$ is in terms of $n$ to narrow down $n$. For a point $p$ there are $n-1$ other points in $S$. These points can be paired by what other point is colinear with $p$, so $n$ is an even number plus one, hence $n$ is odd.
$L_p$ is the set of lines through $p$, so $\lvert L_p\rvert=\frac{n-1}{2}$, as each line passes through 2 other points. Do this for all points in $S$ and add them together to get $\lvert L\rvert=\frac{n(n-1)}{2}$, but we have triple counted every line, so actually, $\lvert L\rvert=\frac{n(n-1)}{6}$
This means $n(n-1)$ is a multiple of 6. $n$ is still odd. Either $n-1=6k$ (hence $n=6k+1$) or $n=6k$, but $n$ is still odd, so that can't be. We also have the possibility of $n$ being divisible by 3 and $n-1$ being even, which is only possible if $n=6k+3$.
That is essentially as far as I've gotten. If thought of it geometrically, limiting $S$ to a convex finite area and a few other things, of which none have worked. All I have as of now is that $n \in \{ 3,7,9,13,15,19,21...\} $
I think what you are looking for is the Sylvester-Gallai theorem, which immediately shows that yes, the only answer is $n=3$. (It states that every finite set of points in the plane has a line either through exactly 2 or through all of them.)