RU-ALLRUSSIAN 1985 Problem 16

$\mathrm{a_1< a_2< \ldots < a_n}$ and $\mathrm{b_1 > b_2 > \ldots > b_n}$ . Taken together the $\mathrm{a_i}$ and $\mathrm{b_i}$ constitute the numbers 1, 2, ..., 2n. Show that $\mathrm{la_1 - b_1| + la_2 - b_2| + \ldots + la_n - b_n| = n^2}$ .