There is a stack of $n$ cards numbered $1, 2, \cdots, n$ 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 single 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.
$$
\begin{array}{l}
\underbrace{1,2, \cdots, k,} k+1, \cdots, 2 k, 2 k+1 \\
\Rightarrow k+1, \cdots, 2 k, 1,2, \cdots, k, 2 k+1 \\
\Rightarrow k+1, \cdots, \underbrace{2 k-1,2,3}_{4}, \cdots, k, 2 k+1,2 k, 1 \\
\Rightarrow k+1, \cdots, 2 k-2,3,4, \cdots, k, 2 k+1,2 k, 2 k-1,2,1 \\
\Rightarrow \ldots \\
\Rightarrow \underbrace{k+1, k, 2 k+1,2 k, \cdots, k+2, k-1, k-2, \cdots, 2,1} \\
\Rightarrow 2 k+1,2 k, \cdots, 2,1 \\
\end{array}
$$

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 now prove that at least $\left[\frac{n}{2}\right]+1$ moves are required.

For any sequence $a_{1}, a_{2}, \cdots, a_{n}$ of $n$ cards, if $a_{i}>a_{i+1}(1 \leqslant i \leqslant n-1)$, then the adjacent pair $\left(a_{i}, a_{1+1}\right)$ 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 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 $f$. Then 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 increase 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 at most 2 inversion pairs. Similarly, it can be proven that for other move types, at most 2 inversion pairs can be increased. Additionally, it is easy to prove that in the first and last moves, each can increase at most 1 inversion pair. 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, at least $\left[\frac{n}{2}\right]+1$ moves are needed to meet the requirement.