Firstly, we notice that we can ignore the orientation of any individual book because if the books return to their original positions in M moves, then another M moves will correct any orientation issues. Assume there are n books, we can show by induction that after 1 round (n moves) the sequence ${\{a_i\}_{i = 1}^n}$, $a_i = i$ which denotes the position of book i from the top, maps to the sequence $\{b_i\}_{i = 1}^n$ where $b_i = n+2-2i, 1 \le i \le \frac{n+1}{2}, b_i = 2i-n-1, \frac{n+2}{2} \le i \le n$. Now we can see that because the mapping is one-to-one, some number of cycles will form were position J will map back to itself in a minimum of K iterations of this mapping. It is also not difficult to see that any position P through which J iterates, also returns to its original position in a minimum of K iterations and that these cycles have no intersection with each other. It follows that if M is the least common multiple of the length of all cycles then M rounds will return all books to their original positions and 2M rounds will definetly return them to their original orientation as well. Q.E.D.