Merlin summons the $n$ knights of Camelot for a conference. Each day, he assigns them to the $n$ seats at the Round Table. From the second day on, any two neighbours may interchange their seats if they were not neighbours on the first day. The knights try to sit in some cyclic order which has already occurred before on an earlier day. If they succeed, then the conference comes to an end when the day is over. What is the maximum number of days for which Merlin can guarantee that the conference will last?
To solve this problem, we need to determine the maximum number of days for which Merlin can guarantee that the conference will last, given the constraints on seat swapping and the requirement for a cyclic order to repeat.

1. **Understanding the Problem:**
   - There are \( n \) knights and \( n \) seats at the Round Table.
   - Each day, the knights are assigned to seats.
   - From the second day onwards, any two neighboring knights can swap seats if they were not neighbors on the first day.
   - The conference ends if the knights sit in a cyclic order that has already occurred on an earlier day.

2. **Initial Observations:**
   - On the first day, the knights are assigned a specific order, say \( A_1, A_2, \ldots, A_n \).
   - On subsequent days, the knights can swap seats with their neighbors under the given condition.
   - We need to find the maximum number of days before a repeated cyclic order occurs.

3. **Permutations and Cyclic Orders:**
   - There are \( n! \) possible permutations of \( n \) knights.
   - However, since the table is round, each permutation can be rotated \( n \) ways, reducing the number of unique cyclic orders to \( \frac{n!}{n} = (n-1)! \).

4. **Swapping Constraints:**
   - From the second day onwards, knights can only swap seats with their neighbors if they were not neighbors on the first day.
   - This constraint limits the number of possible new arrangements each day.

5. **Counting Unique Arrangements:**
   - On the first day, there is 1 unique arrangement.
   - On the second day, the knights can swap seats with their neighbors, but the number of valid swaps is limited by the first day's arrangement.
   - We need to determine how many unique arrangements can be generated under these constraints before a repeat occurs.

6. **Maximum Number of Days:**
   - The maximum number of days is determined by the number of unique cyclic orders that can be generated before a repeat occurs.
   - Given the constraints, the maximum number of unique cyclic orders is \( (n-1)! \).

7. **Conclusion:**
   - The maximum number of days for which Merlin can guarantee that the conference will last is \( (n-1)! \).

The final answer is \( \boxed{ (n-1)! } \).