In a building there are 6018 desks in 2007 rooms, and in every room there is at least one desk. Every room can be cleared dividing the desks in the oher rooms such that in every room is the same number of desks. Find out what methods can be used for dividing the desks initially.
1. **Determine the maximum number of desks in a room:**
   Let \( n \) be the maximum number of desks in a room. We know that there are 6018 desks in 2007 rooms, and each room has at least one desk. Therefore, we have:
   \[
   1 \leq n \leq \left\lfloor \frac{6018}{2007} \right\rfloor = 3
   \]
   Since \( n \) must be an integer, we have \( n \geq 3 \). If \( n \leq 2 \), then:
   \[
   6018 \leq 2007n \leq 4014
   \]
   which is false. Therefore, \( n \geq 3 \).

2. **Check if \( n \geq 4 \) is possible:**
   If \( n \geq 4 \), then choose another room to be cleared. After redistribution, in all "full" (not cleared) rooms, we have at least \( n \) desks. Thus:
   \[
   6018 \geq 2006n \geq 8024
   \]
   which is false. Therefore, \( n = 3 \).

3. **Analyze the distribution of desks:**
   We observe that there are missing 3 desks so that in all rooms there would be 3 desks. We take two cases:

   - **Case 1: There is a room with 1 desk.**
     The distribution is \( 1, 2, 3, 3, 3, \ldots, 3 \) (and permutations). This kind of distribution satisfies all the tasks because we can do moves like this:
     - If we clear the room with 1 desk, then move this desk into the room with 2 desks.
     - If we clear the room with 2 desks, then move these 2 desks into the room with 1 desk.
     - If we clear any of the rooms with 3 desks, then move 2 of these desks into the room with 1 desk and the other 1 desk into the room with 2 desks.

   - **Case 2: There is no room with 1 desk.**
     The distribution is \( 2, 2, 2, 3, 3, \ldots, 3 \) (and permutations). This kind of distribution satisfies all the tasks because we can do moves like this:
     - If we clear any of the rooms with 2 desks, then take the other 2 rooms with 2 desks and put in each of them 1 desk.
     - If we clear any of the rooms with 3 desks, then take all the rooms with 2 desks and put in each of them 1 desk.

4. **Conclusion:**
   The possible initial distributions of desks are:
   \[
   1, 2, 3, 3, 3, \ldots, 3 \quad \text{and} \quad 2, 2, 2, 3, 3, \ldots, 3 \quad \text{(and permutations)}
   \]

The final answer is \( \boxed{ 1, 2, 3, 3, 3, \ldots, 3 } \) and \( 2, 2, 2, 3, 3, \ldots, 3 \) (and permutations).