Find an example of a set of points of the plane such that there exist at least $2^{n-2}$ good partitions. If a set of $n$ points of the plane can be divided into two subsets $\{ A_1, A_2, \dots,  A_k\}$ and $\{ B_1, B_2, \dots, B_{n-k}\}$  so that there is a point $M$ of the plane such that
$$ MA_1+ MA_2+ \dots+MA_k= MB_1+ MB_2+ \dots+MB_{n-k},$$
then we shall say that there exists a good partition of the set of these $n$  points of the plane. 
Prove that for all $n \ge 2$ there is a set of $n$ points of the plane such that the number of good partition of this set is not less than $2^{n-2}$.
Partitions in which two sets $A$ and $B$ change between themselves are the same partitions.
My work. I tried to place points at the vertices of a regular polygon or on a line at the same distance from each other. But it is difficult to prove the problem for these cases.
 A: Put $n-1$ points in a ball of radius $\epsilon$ and one point that is at distance $R$ from the origin. Make $\epsilon = R/(1000n)$. Pick any point in the circle and the point that is far away. Call these points special points. Now partition the other $n-2$ points into two sets arbitrarily. Call these partitions set $A$ and $B$. 
Now extend the partition by adding the special point that is inside the circle in way that the two sets have different sizes. We can assume that $|A| > |B|$ after adding in this one special point. Now assign the other special point that is far away to set $B$. This gives us $2^{n-2}$ possible partitions. We show that each partition is good.
Let $f(M) = \sum_{A_i \in A} MA_i - \sum_{B_i \in B} MB_i$. Note that $f$ is a continuous function in $M$. Now let $M_1$ be the origin of the circle. We can compute that
$$ f(M_1) \le \frac{Rn}{500n} -R < 0.$$
Let $M_2$ be the special point that is very far away from the circle. We have
\begin{align*}
f(M_2) &\ge (R-\epsilon)|A|-(R+\epsilon)(|B|-1)  \\
       &\ge R(|A|-(|B|-1)) - \epsilon n \\
       &\ge R-\epsilon n > 0
\end{align*}
where we have used the fact that $|A| > |B|-1$ due to the way we added the special points.
Then by the intermediate value theorem, there exists some point in the plane such that $f = 0$, as desired.
