Median vs. Mean Problem:
Consider the following model:
$y_i = \mu + \epsilon_i$, $i = 1,...,n$
Let the mean $\mu$ be estimated by minimizing the criterion $\sum|\mu - y_i|$ over $\mu$.
Show that $m = $median($y_1,y_2,...,y_n$) is optimal for this criterion.
Distinguish the case $n$ is odd and $n$ is even.

My idea to approach the problem:
$1)$ Rewrite the criterion such that no absolute signs are needed (with indicator function for example).
$2)$ Determine the first order condition to recognize the median.
$3)$ Order data to distinguish $n$ odd or $n$ even

However, I dont't know how to to the steps of my apporach in a mathematical/statistical way.
Could anyone please help me?

Your approach looks good.
$$
\sum |\mu - y_i| = \sum_{y_i< \mu} [\mu - y_i]+ \sum_{y_i> \mu} [y_i - \mu]
$$
now suppose that $y_1< y_2 <\cdots < y_n$, and let $k(\mu)$ the unique $k$ such as
 $y_k <  \mu < y_{k+1}$. First order conditions are enough to find the solution because the function is convex, and:
$$
\sum |\mu - y_i| = \sum_{i=1}^{k(\mu)} [\mu - y_i]+ 
\sum_{i=k(\mu)+1}^n [y_i - \mu]\\
0 = \frac{d}{d\mu} \sum |\mu - y_i| = k(\mu) - (n-k(\mu)-1)
$$


*

*if $n$ is odd, any $\mu\in [y_{{n-1}/2}, y_{{n+1}/2}]$ minimizes the function, in particular the median $\frac 12[y_{{n-1}/2}+ y_{{n+1}/2}]$

*if $n$ is even, the derivative is never $=0$ but changes its sign when $\mu = y_{n/2}$. The median is the unique minimizer.

