Here is a solution. First of all, WLOG assume that the rectangles all have sides parallel to one another. Indeed, if this were not the case it suffices to consider each "orientation" separately and use the fact that $\sum_{i = 1}^{n} 4 \sqrt{n_i} > 4 \sqrt{n}$ for positive integers $n_1, n_2, \cdots, n_k$ summing to $n$. With this assumption, toss the rectangles in the Cartesian coordinate plane so that all sides are parallel to the $x-$ or $y-$axis. We call a right angle "Type One" if it contains the point $(x, y)$ whenever $x, y$ are sufficiently large. In other words, it can be translated in a way that it contains all points in the first quadrant. Similarly define "Type Two," "Type Three," and "Type Four" for the second, third, and fourth quadrant respectively. Let $a, b, c, d$ be the number of Type One, Type Two, Type Three, and Type Four right angles respectively. Lemma. $ac \ge n$. Proof. Consider the mapping which takes each rectangle to the Type One and Type Three right angles it contains (i.e. the right angles at its bottom-left and top-right corners). This is clearly an injection from our $n$ rectangles to the pairs of Type One and Type Three right angles and so $n \le ac.$ $\blacksquare$ This implies that $a + c \ge 2 \sqrt{n}$ by AM-GM. Analogously we can show that $b + d \ge 2 \sqrt{n}$ and so summing gives $a+ b + c + d \ge 4 \sqrt{n},$ which yields the result. $\square$

Hi everyone! I just can't resist drawing the curtain. This is Problem 14 of Iran TST 2013.