Solution:

First of all, let's make the whole picture easier to handle: We can split the rectangles into groups such that in each group, all sides of rectangles are parallel or perpendicular to each other. We also choose these groups to be maximal, in particular: For any two rectangles of different groups, they don't have parallel sides. We observe that this way, no two right angles of different groups can be the same, so we may count the right angles in each group and sum it up in the end. Say there are $k$ different groups and denote the number of rectangles in the $i$-th group by $n_{i}$. We can choose a coordinate system that has axes parallel to the rectangles in the $i$-th group and from now on only consider rectangles in this group.

We can now introduce the following variables:
- $A_{i}$ : number of distinct right angles that form a top left corner
- $B_{i}$ : number of distinct right angles that form a top right corner
- $C_{i}$ : number of distinct right angles that form a bottom left corner
- $D_{i}$ : number of distinct right angles that form a bottom right corner

Since every rectangle in the $i$-th group is uniquely determined by one top left and one bottom right corner and for any two distinct rectangles, the combination of top left and bottom right corner must be different, we get the estimation
$$A_{i} \cdot D_{i} \geq n_{i}$$
The same argument for top right corners and bottom left corners yields
$$B_{i} \cdot C_{i} \geq n_{i}$$
Now, by applying AM-GM to both expressions on the left hand sides, we get
$$\begin{gathered} \left(\frac{A_{i}+D_{i}}{2}\right)^{2} \geq A_{i} \cdot D_{i} \geq n_{i} \Rightarrow A_{i}+D_{i} \geq 2 \sqrt{n_{i}} \\ \left(\frac{B_{i}+C_{i}}{2}\right)^{2} \geq C_{i} \cdot B_{i} \geq n_{i} \Rightarrow B_{i}+C_{i} \geq 2 \sqrt{n_{i}} \\ \Rightarrow A_{i}+B_{i}+C_{i}+D_{i} \geq 4 \sqrt{n_{i}} \end{gathered}$$
In other words, the number of distinct right angles in the $i$-th group is greater or equal than $4 \sqrt{n_{i}}$. Summing over all groups, we get the inequality:
$$\sum_{i=1}^{k}\left(A_{i}+B_{i}+C_{i}+D_{i}\right) \geq \sum_{i=1}^{k} 4 \sqrt{n_{i}}$$
On the left hand side, we have the total number of distinct right angles. So it remains to show $\sum_{i=1}^{k} 4 \sqrt{n_{i}} \geq 4 \sqrt{n}$, which is obvious by
$$\left(\sum_{i=1}^{k} \sqrt{n_{i}}\right)^{2} \geq \sum_{i=1}^{k}{\sqrt{n_{i}}}^{2}=\sum_{i=1}^{k} n_{i}=n$$