Proving that median of list $[x_1,x_2,...,x_n]$ minimises the sum $\sum_{i=1}^{i=n} |x_i-m|$ where $m$ is some number The problem is in the title. Here is a detailed description:
Let's say we have list $[x_i]_{i=1}^{i=n}$ where $x_i\in\Bbb{N}$. I want to pick such $m\in\Bbb{N}$ which minimises the sum $\sum_{i=1}^{i=n} |x_i-m|$.
In my computer science class we had a problem which was equivalent to the upper one. And apparently the answer was that the $m$ we should choose is the median of the list. But no one has proven it, and I have a hard time seeing why that is the case and not for example arithmetic mean of the list. 
What strategy should I use in order to prove that? 

Hint: The derivative of the sum with respect to $m$ is $$\frac{\partial }{\partial m}\sum_{i=1}^{n}|m-x_i|=\sum_{i=1}^n{}\frac{m-x_i}{|m-x_i|}$$ where $$\frac{m-x_i}{|m-x_i|}=\begin{cases}\phantom{-}1, & m>x_i \\[0.2cm]-1,& m<x_i\end{cases}$$ and therefore the sum (that appears in the derivative) is equal to zero for $m_0$ such that half of the observations are positive (+1) and half of the observations are negative (-1). Thus the median $m_0$ is a critical point. Due to convexity of the absolute value this is a minimum. 
(Of course if there exists $x_i=m_0$ then the function is not differentiable at that point, but the reasoning for obtaining a minimum applies). 
