Proof that median of an array is the number that minimizes the sum of manhattan distance to all points
marks the set of numbers P={P1,P2,P3,..,Pn}

need to find number X:

|P1-X|+|P2-X|+...+|Pn-X|=D, which minimizes D.

the fact that X is the median can be shown using induction on the number of numbers n in P.

Base: n=1: trivial to see that the median is P1 and |P1-P1|=D=0 is minimal (D>=0).

now suppose that this holds for all P of size |P|=n, we need to show that this holds for |P|=n+1.

In other words:

given that |P1-X|+|P2-X|+...+|Pn-X|=D is minimal when X=Pm is the median of P={P1,..,Pn} show that |P1-X|+|P2-X|+...+|Pn-X|=D is minimal when X=Pm' is the median of P={P1,..P(n+1)). 
notice that there are two cases m'=m+1 (n is even) or m'=m (n is odd).

I suggest drawing a simple example using n=3 and n+1=4 on an axis. Can you see why this must hold?