Let $\angle CAB = 2\alpha$, $\angle ABC = 2\beta$ and $\angle BCA = 2\gamma$. Notice that $\alpha+\beta+\gamma = 90^\circ$, and if $I$ is the incenter of $\triangle ABC$ we can compute $\angle AIB$, $\angle BIC$, $\angle CIA$ in terms of these variables.

Fact 1: The number of $60^\circ$ angles in any step of the process is equal to the number of $120^\circ$ angles in the next step.
Proof: It suffices to prove that each $60^\circ$ angle generates a $120^\circ$ angle, and that every $120^\circ$ angle is generated by a $60^\circ$ angle.
The first statement is clear: if, say, $2\alpha = 60^\circ$, then $90^\circ + \alpha = 120^\circ$. For the second statement, observe that the only way we can obtain a $120^\circ$ angle is if one of the angles at I measures $120^\circ$, because those are the only obtuse angles generated. For this to be true, one of our variables must be equal to $30^\circ$, which in turn means one of the angles of ABC must be $60^\circ$.
Fact 2: The number of $120^\circ$ angles in any step of the process is half the number of $60^\circ$ angles in the next step.
Proof: When performing a step, every $120^\circ$ angle gets divided into two $60^\circ$ angles. Moreover, this is the only way of obtaining a $60^\circ$ angle. $\square$
Using these two facts we can readily complete the following table, which shows the answer is $32$.