$ 2^n $ coins are given to a couple of kids. Interchange of the coins occurs when some of the kids has at least half of all the coins. Then from the coins of one of those kids to the all other kids are given that much coins as the kid already had. In case when all the coins are at one kid there is no possibility for interchange. What is the greatest possible number of consecutive interchanges? ($ n $ is natural number)
1. **Base Case:**
   - For \( n = 1 \), we have \( 2^1 = 2 \) coins. Suppose one kid has 1 coin and the other kid has 1 coin. The condition for interchange is that one of the kids must have at least half of all the coins, which is 1 coin in this case. Since both kids have 1 coin, an interchange can occur. After the interchange, the distribution of coins remains the same. Thus, the number of interchanges is 1, which is equal to \( n \).

2. **Inductive Step:**
   - Assume that for \( n = k \), after the \( k \)-th interchange, the number of coins each kid has is divisible by \( 2^k \). We need to show that for \( n = k+1 \), after the \( (k+1) \)-th interchange, the number of coins each kid has is divisible by \( 2^{k+1} \).

3. **Inductive Hypothesis:**
   - Suppose after \( k \) interchanges, the number of coins each kid has is divisible by \( 2^k \). Let the total number of coins be \( 2^{k+1} \). Initially, one kid has \( 2^{k+1} - 1 \) coins and the other kid has 1 coin.

4. **First Interchange:**
   - The kid with \( 2^{k+1} - 1 \) coins gives \( 2^{k+1} - 1 \) coins to the other kid. After the first interchange, the distribution of coins is:
     \[
     \left( 2^{k+1} - 1 - (2^{k+1} - 1), 1 + (2^{k+1} - 1) \right) = (0, 2^{k+1})
     \]
   - Now, the kid with 0 coins cannot give any coins, so the interchange stops.

5. **General Case:**
   - For any \( n \), the number of interchanges is limited by the fact that after each interchange, the number of coins each kid has is divisible by \( 2^k \). Therefore, the maximum number of interchanges is \( n \).

6. **Example for \( n = 3 \):**
   - Suppose we have \( 2^3 = 8 \) coins. Initially, one kid has 7 coins and the other has 1 coin.
     - After the 1st interchange: (7, 1) → (0, 8)
     - After the 2nd interchange: (0, 8) → (8, 0)
     - After the 3rd interchange: (8, 0) → (0, 8)
   - The process stops here as the distribution repeats.

Conclusion:
The greatest possible number of consecutive interchanges is \( n \).

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