Example 2: A rope of length 2009 is operated as follows: first, it is divided into two ropes of positive integer lengths, and the lengths of the two ropes are recorded, then the above operation is repeated on one of the ropes, ... until 2009 ropes of length 1 are obtained. If the lengths of the two ropes obtained in a certain operation are not equal, then this operation is called "good".
(1) Find the maximum and minimum values of the number of good operations;
(2) Prove: In all operation processes where the number of good operations reaches the minimum value, the number of different lengths of ropes recorded is the same.


【Analysis】(1) It is easy to know that a rope of length 2 can only be divided into two segments of length 1, meaning the last operation on the rope must not be a good operation. A rope of length 2009 can be divided into 2009 segments of length 1 after exactly 2008 operations, so the number of good operations is no more than
$2008-1=2007$.
Therefore, the maximum number of good operations is 2007.
The following is such an operation: each time, cut a segment of length 1, then the first 2007 operations are all good operations.

Generalize the problem: For a rope of length $n$, let $f(n)$ be the minimum number of good operations.

By conducting mathematical experiments with simple positive integers $n$ and summarizing the experience, the following two lemmas can be discovered.
Lemma 1: $f(n)=0 \Leftrightarrow n=2^{k}$.
Lemma 2: Suppose $n=2^{k_{1}}+2^{k_{2}}+\cdots+2^{k_{1}}\left(k_{1}>k_{2}\right.$ $\left.>\cdots>k_{l} \geqslant 0\right)$. Let $l=S_{2}(n) \triangleq$ the sum of the digits of $n$ in binary. Then $f(n)=S_{2}(n)-1$.

Proof of Lemma 2: Since after $l-1$ good operations, we can obtain $2^{k_{1}}, 2^{k_{2}}, \cdots, 2^{k_{1}}$, then
$f(n) \leqslant l-1$.
We will prove by mathematical induction:
$f(n) \geqslant S_{2}(n)-1$.
When $n=1$, it is Lemma 1, and the conclusion is obviously true.
Assume that for $n \leqslant k$, $f(n) \geqslant S_{2}(n)-1$.
When $n=k+1$, by the definition of $f$, there exists an operation process containing $f(k+1)$ good operations and several non-good operations, which cuts a rope of length $k+1$ into $k+1$ segments of length 1. If the first operation in the above process is a good operation, cutting a rope of length $k+1$ into segments of lengths $a$ and $b$, i.e., $k+1=a+b, a \neq b$, then
$$
\begin{array}{l}
f(k+1)=f(a)+f(b)+1 \\
\geqslant\left(S_{2}(a)-1\right)+\left(S_{2}(b)-1\right)+1 \\
=S_{2}(a)+S_{2}(b)-1 \\
\geqslant S_{2}(a+b)-1=S_{2}(k+1)-1,
\end{array}
$$

The equality holds if and only if there is no carry in the binary addition of $a+b$.

If the first operation in the above process is not a good operation, i.e., $k+1=2a$, then
$$
f(k+1)=2 f(a) \geqslant 2\left(S_{2}(a)-1\right) .
$$

Since $S_{2}(2a) \leqslant S_{2}(a)+S_{2}(a)-1$, the equality holds if and only if $S_{2}(a)=1$, i.e.,
$$
2 S_{2}(a)-2 \geqslant S_{2}(2a)-1 .
$$

Using this result, we get
$$
\begin{array}{l}
f(k+1) \geqslant 2\left(S_{2}(a)-1\right) \\
\geqslant S_{2}(2a)-1=S_{2}(k+1)-1,
\end{array}
$$

The conclusion holds, and the equality holds if and only if $S_{2}(a)=1$.
In summary, Lemma 2 is proved.
Therefore, using the binary representation of 2009, we get
$$
f(2009)=S_{2}(2009)-1=7 \text {. }
$$
(2) If the binary representation of $n$ is
$$
n=2^{k_{1}}+2^{k_{2}}+\cdots+2^{k_{i}} \text {, }
$$

we can set the set $M=\left\{k_{1}, k_{2}, \cdots, k_{l}\right\}$.
If $l \geqslant 2$, then in any operation process where the number of good operations reaches the minimum, the first operation is a good operation, i.e., there exists a non-empty proper subset $L$ of $M$ such that the rope of length $n$ is cut into segments of lengths $a$ and $b$, where $a=\sum_{k \in L} 2^{k}, b=\sum_{k \in M-L} 2^{k}$. Performing $l-1$ good operations in succession produces $l$ different lengths $2^{k_{1}}, 2^{k_{2}}, \cdots, 2^{k_{i}}$.

Then, the non-good operations produce lengths $2^{k_{1}-1}, 2^{k_{1}-2}, \cdots, 2,2^{0}$, and so on. Therefore, the number of different lengths recorded is $l-1+k_{1}$.

In solving this problem, it is found through mathematical experiments that the minimum number of good operations $f(n)$ is related to the binary representation of the positive integer $n$, and a basic result about the sum of digits is used:
$$
S_{p}(a+b)=S_{p}(a)+S_{p}(b)-k(p-1),
$$

where $k$ is the number of carries in the addition of $a+b$ in base $p$.