9.6 There is a red card box and $k$ blue card boxes $(k>1)$, and a deck of cards, totaling $2n$ cards, which are numbered from 1 to $2n$. Initially, this deck of cards is stacked in the red card box in any order. From any card box, the top card can be taken out, either placed in an empty box, or placed on top of a card with a number exactly one less. For what maximum $n$ can all the cards be moved to one of the blue card boxes?
$9.6 \quad n=k-1$.
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 first with odd-numbered cards (in any order), then the card numbered $2n$, and finally 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), and then, 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 at this time.

Assume $n < k$, we will prove that it is always possible to move the cards as required.
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 in the red box to its "own" blue box. If at some step it cannot be done, it must be that 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, place the card numbered $2i$ in the "free box", then move the card numbered $2i-1$ on top of it, and change their original blue box to a "free box". This way, each pair of cards can be moved into a blue box, with the smaller number on top. At this point, using the "free box", all cards can be moved into one blue box.