Example G: There is a stack of $n$ cards numbered from $1, 2, \cdots, n$ from top to bottom. Each time, a small stack of cards (which can be just one card) can be taken out from this stack, and inserted as a whole without changing the order into any position of the original stack (it can be inserted between two cards, or placed on top of the first card or below the $n$-th card). Such an operation is called a move. Question: What is the minimum number of moves required to completely reverse the order of the stack, so that the $n$ cards are numbered from top to bottom as $n$, $n-1, \cdots, 2, 1$?
If $n=2 k+1$, then by performing $k+1$ moves as shown in the table below, the order can be completely reversed.
$$
\begin{array}{l}
\Rightarrow 2 k+1,2 k, \cdots, 2,1 \\
\end{array}
$$

If $n=2 k$, by removing the card numbered $2 k+1$ from the table above, it is clear that $k+1$ moves are sufficient. We will now prove that the minimum number of moves required is $\left[\frac{n}{2}\right]+1$.

For any sequence of $n$ cards $a_{1}, a_{2}, \cdots, a_{n}$, if $a_{i}>a_{i+1}(1 \leqslant i \leqslant n-1)$, then the adjacent pair $\left(a_{i}, a_{i+1}\right)$ is called an inversion pair. In the initial state, there are no inversion pairs, and in the final state, there should be $n-1$ inversion pairs. We will now examine the effect of each move on the number of inversion pairs.

Suppose a move is made from “$\cdots a b \cdots c d \cdots e f$” by extracting “$b \cdots c$” and inserting it between $e$ and $f$. The changes in all adjacent pairs are: removing $(a, b),(c, d),(e, f)$, and adding $(a, d),(e, b),(c, f)$. We will prove that it is impossible to add 3 inversion pairs.

This is because, if $a>d, e>b, c>f$, then $a<b<e<f<c<d<a$ leads to a contradiction! Therefore, each such move can increase the number of inversion pairs by at most 2. Similarly, it can be proven that for other move types, the number of inversion pairs can also increase by at most 2. Additionally, it is easy to prove that in the first and last moves, each can increase the number of inversion pairs by at most 1. Therefore, to increase the number of inversion pairs from 0 to $n-1$, at least $\left[\frac{n}{2}\right]+1$ moves are required.
In summary, the minimum number of moves required is $\left[\frac{n}{2}\right]+1$.

/Comment: Construction and proof are common methods for solving such problems. In this problem, using a semi-invariant (the number of inversion pairs) to prove the lower bound of the number of moves is a simple and effective method.