We are given one red and $k>1$ blue cells, and a pack of $2n$ cards, enumerated by the numbers from $1$ to $2n$.  Initially, the pack is situated on the red cell and arranged in an arbitrary order.  In each move, we are allowed to take the top card from one of the cells and place it either onto the top of another cell on which the number on the top card is greater by $1$, or onto an empty cell.  Given $k$, what is the maximal $n$ for which it is always possible to move all the cards onto a blue cell?
1. **Counterexample for \( n = k \)**:
   - Suppose the red cell has cards in the order \( 1, 3, \ldots, 2k-1, 2k \), and then the rest in arbitrary order.
   - After \( k \) moves, all \( k \) cells will be filled with the odd numbers from \( 1 \) to \( 2k-1 \), since none can be placed on each other.
   - Then \( 2k \) can't go anywhere because there are no empty cells and no cell with a top card \( 2k-1 \).

2. **Proof that \( n = k-1 \) is achievable**:
   - **Definition of a "good" position**: A position is "good" if it has the cards from \( 1 \) to \( k-1 \) in one cell and the cards from \( k \) to \( 2k-2 \) in another cell.
   
   - **Step 1: A good position leads to a win**:
     - Put \( 1, 2, \ldots, k-2 \) successively in the \( k-2 \) empty cells.
     - Now put \( k-1 \) on top of the stack with \( k \) at the top, then \( k-2 \) and continue until \( 1 \) is placed and we have one large stack.

   - **Step 2: Induction on \( k \)**:
     - **Base case**: For \( k=2 \) and \( n=1 \), it is easy because we just put \( 1 \) and \( 2 \) in separate cells regardless of their order.
     - **Inductive step**: Suppose that we can achieve a good position with \( n=k-1 \) in \( k \) cells. We will show it is true for \( k+1 \) cells, \( n=k \).
       - Choose one cell and set it aside for the cards with \( 2k-1 \) and \( 2k \).
       - Use the process that gets a good position for \( n=k-1 \) in the remaining \( k \) cells.

   - **Step 3: Achieving a stack with \( 2k-1 \) and \( 2k \) in one cell**:
     - In every case, it is obviously true except if \( 2k-1 \) comes before \( 2k \) and if when \( 2k \) comes all the \( k \) cells are full.
     - By the pigeonhole principle, there must be two top cards from one of the pairs \( (1,2), \ldots, (2k-3, 2k-2) \), so two of the top cards must differ by \( 1 \), say \( a \) and \( a+1 \).
     - \( a \) must have been the only card in its stack, so we can move \( a \) on top of \( a+1 \).
     - Then we place \( 2k \) in the cell where \( a \) was, put \( 2k-1 \) on top of it, then return \( a \) to the cell where \( 2k-1 \) was.
     - So the process for \( n=k-1 \) has not been disturbed.

   - **Step 4: Achieving a good position**:
     - Place the cards from \( k \) to \( 2k-3 \) in the \( k-2 \) free cells.
     - Then successively place \( 2k-2 \) to \( k+1 \) onto the stack with \( 2k-1 \) and \( 2k \).
     - Then place cards from \( 1 \) to \( k-2 \) in the \( k-2 \) free cells.
     - Then successively place cards from \( k-1 \) to \( 1 \) onto the cell with \( k \).
     - Then we have achieved a good position, which completes the induction.

\(\blacksquare\)

The final answer is \( \boxed{ n = k-1 } \)