An algebra problem by U Z

The possible medians are the numbers between $3$ and $9$ inclusive (see below for some possible ways to do this), for a total of $\boxed{7}$ different possible values:

$\color{#3D99F6}-1\color{#333333}, \color{#3D99F6}0\color{#333333}, \color{#3D99F6}1\color{#333333}, 2, \boxed{3}, 4, 6, 9, 14$

$\color{#3D99F6}0\color{#333333}, \color{#3D99F6}1\color{#333333}, 2, 3, \boxed{4}, 6, 9, 14, \color{#3D99F6}15\color{#333333}$

$\color{#3D99F6}1\color{#333333}, 2, 3, 4, \boxed{\color{#3D99F6}5\color{#333333}}, 6, 9, 14, \color{#3D99F6}15\color{#333333}$

$\color{#3D99F6}1\color{#333333}, 2, 3, 4, \boxed{6}, 9, 14, \color{#3D99F6}15\color{#333333}, \color{#3D99F6}16\color{#333333}$

$2, 3, 4, 6, \boxed{\color{#3D99F6}7\color{#333333}}, 9, 14, \color{#3D99F6}15\color{#333333}, \color{#3D99F6}16\color{#333333}$

$2, 3, 4, 6, \boxed{\color{#3D99F6}8\color{#333333}}, 9, 14, \color{#3D99F6}15\color{#333333}, \color{#3D99F6}16\color{#333333}$

$2, 3, 4, 6, \boxed{9}, 14, \color{#3D99F6}15\color{#333333}, \color{#3D99F6}16\color{#333333}, \color{#3D99F6}17\color{#333333}$

Numbers less than $3$ and greater than $9$ cannot be the medians because we can only add three numbers to the list.