Example 1. There are several different books, which are initially numbered as No. 1, No. 2, $\cdots$, up to No. $\mathrm{n}$. Later, they are renumbered again. This time, the numbering is not entirely the same, so each book has two numbers: the new number and the old number. If the two numbers of each book are different, then the same number may correspond to two different books: one book is numbered as such according to the new number, and another book is numbered as such according to the old number. This can easily cause confusion. Now, consider placing these books in separate piles so that in each pile, one number corresponds to only one book. Then, how many piles are needed to achieve this?

Our answer is: Just 3 piles are enough.
The 1st book can be placed in any pile at will. Suppose that the first $\mathrm{k}-1$ books have been successfully sorted into piles, we now consider the placement of the $\mathrm{k}$-th book: the old number in this book may match the new number of one of the $\mathrm{k}-1$ books already sorted, in which case this book cannot be placed in that pile; the new number of this book may also match the old number of one of the books already sorted, in which case this book cannot be placed in that pile either; apart from these, no other duplicate numbers will occur, so at most, it cannot be placed in two piles, meaning there is always one pile available for it. Therefore, the $\mathrm{k}$-th book can be successfully placed. By mathematical induction: these $\mathrm{n}$ books can be divided into three piles, ensuring that in each pile, one number corresponds to one book.