Example 7 (Problem from the 28th Russian Mathematical Olympiad): There is a red card box and $k$ blue card boxes ($k>1$), and a deck of cards, totaling $2n$ cards, numbered from 1 to $2n$. Initially, these cards are stacked in the red card box in any order. From any card box, the top card can be taken out, placed in an empty box, or placed on top of a card with a number exactly one greater. For what largest $n$ can all the cards be moved to one of the blue card boxes through these operations?
First, we prove that for larger $n$, it is not necessarily possible. Suppose at the beginning, the cards in the red box are placed from top to bottom with odd-numbered cards first (in any order), followed by the card numbered $2n$, and then the remaining even-numbered cards (in any order). In this case, the first $k$ steps are uniquely determined (odd-numbered cards are successively placed in the empty blue box). Next, if $n > k$, no further moves can be made; if $n = k$, only the card numbered $2n-1$ can be returned to the red box, which is equivalent to no movement. Therefore, it is impossible to move the cards as required in this scenario.

Assume $n < k$, we will prove that it is always possible to move the cards as required.

We pair the cards as follows:
$$
(1,2),(3,4), \cdots,(2n-1,2n),
$$

and assign each pair to an empty blue box. At this point, there is at least one blue box without a pair assigned (called a "free box"). We sequentially move the topmost card from the red box to its "own" blue box. If at some step it is not possible to proceed, it must be because a card numbered $2i-1$ has already entered its "own" blue box, preventing the card numbered $2i$ from entering its "own" blue box. In this case, we place the card numbered $2i$ in the "free box," then move the card numbered $2i-1$ on top of it, and reclassify their original blue box as a "free box." This way, we can move each pair of cards into a blue box with the smaller number on top. At this point, using the "free box," we can move all the cards into one blue box.

Therefore, the maximum $n = k-1$.