Prove that there is only a single point of minimum distance for $N>3$ points We want to mimimize the sum of distances from $n$ distinct points. Prove that there exists only one such point for $n>3$ if all the $n$ points lie on a single plane (and not on a single line)
The problem seems quite tough, but might possess an elegant solution. I tried creating two PDE's (partial differential equations) for $x$ and $y$ coordinates and couldn't find anything fruitful. Might be something towards vectors and their sum. 
 A: Hints & References
You are seeking to prove that the Geometric median is unique. This holds true only if the points are not colinear, an assumption that should be added at the description.
This paper-"The multivariate L1-median and associated data depth"-presents a generalized approach. 
This answer on-"How to find out Geometric Median" on stackoverflow provides an algorithm for finding such a point. 
Also this post-"Geometric median (or Fermat-Weber problem), including continuous case"-here on MSE might interest you-
A: lemma:suppose $A,B,C$ are three distinic points on plane , and $M$ is the middle point of $BC$, then $AM \le {{AB+AC} \over 2} $
proof: 
if three points $A,B,C$ lie on a line it is clear the condtion is true, now consider the case which three point create a triangle , suppose $A_1$ be the symmetry point of $A$ into $M$ then quadrilateral $ABA_1C$ diameters , cuting each other in to half , so it is a parallelogram.please note that $AA_1=2AM$ and $AB=CA_1$ now accordin to the triangle inequality in the $AA_1C$ we have $AA_1 <AC+CA_1$ thus $2AM<AB+AC$ and it is proved.
now lets solve problem  using this lemma:
suppose ${A_1,A_2,A_3,....,A_n}$ are $n$ distinic point and suppose there exist two point that has minimumthe sum of distances from $n$ distinct points, call them $X$ and $Y$.
call the middle point of $X$ and $Y$ the $M$,now we are using above lemma in $A_1XY$ and $A_2XY$ and ... $A_nXY$ (please note that even if three point lies on a line then lemma is true):
$A_1M \le {{A_1X+A_1Y}\over 2}$
$A_2M \le {{A_2X+A_2Y}\over 2}$
.....
$A_nM \le {{A_nX+A_nY}\over 2}$
add all of them ,$A_1M+A_2M+...+A_nM \le { {A_1X+...+A_nX+A_1Y+...+A_nY} \over 2}$ but since all the points is not on a single line so at least one of the inequalitys is strict inequality thus $A_1M+A_2M+...+A_nM < { {A_1X+...+A_nX+A_1Y+...+A_nY} \over 2}$, and since $X$ and $Y$ has minimum sum,call it H, so $A_1M+A_2M+...+A_nM < H={2H\over 2}={ {A_1X+...+A_nX+A_1Y+...+A_nY} \over 2}$ so the point $M$ is the desired point which is contradiction.(note you should add all the point is not on a single line other wisee there exist two points)
A: Here is your sum
$$
S(r) = \sum_{k=1}^n \lVert r - r_k \rVert
$$
where $r, r_k \in \mathbb{R}^N$. The $i$-th component of the $r$-gradient of $S$ in the Euclidean norm is
\begin{align}
\partial_i S(r) 
&= \partial_i \sum_{k=1}^n \lVert r - r_k \rVert \\
&= \partial_i \sum_{k=1}^n \sqrt{\sum_{j=1}^N \left(x_j - x_{k,j} \right)^2} \\
&= \sum_{k=1}^n \frac{1}{2}\lVert r - r_k \rVert^{-1} \sum_{j=1}^N 2(x_j-x_{j,k})\delta_{ij} \\
&= \sum_{k=1}^n \frac{x_i-x_{k,i}}{\lVert r - r_k \rVert}
\end{align}
So critical points $r$ of $S$ (which are candidates for extrema) fulfill the condition
$$
\sum_{k=1}^n \frac{r-r_k}{\lVert r - r_k \rVert} = 0
$$
This seems to be the characterization of the geometric median (Wikipedia article).
The Wikipedia article links to Vardi & Zhang (2000) for a uniqueness proof, which was already cited by MathematicianByMistake. 
A: Suppose the minimum sum occurs at two points, $P$ and $Q$.  Let $M$ be the midpoint of the line segment $PQ$. For all points not on the line $PQ$, the distance to $M$ is strictly less than the average of the distance to $P$ and $Q$.  Since it's assumed that not all $n$ points are on a single line, the sum of their distances to $M$ is strictly less than the average of their sums to $P$ and $Q$ -- which is to say, strictly less than the minimum sum.  That's a contradiction, hence the mimimum sum occurs at a unique point.
Remark:  As Nick Pavlov points out, this answer essentially duplicates Dexpectra's. That answer contains a nice proof of the key observation, that the distance to $M$ is less than the average of the distances to $P$ and $Q$ (by viewing $M$ as the center of a parallelogram with $P$ and $Q$ as oppositve vertices).
