The natural number $n>1$ is called “heavy”, if it is coprime with the sum of its divisors. What’s the maximal number of consecutive “heavy” numbers?
1. **Understanding the definition of "heavy" numbers:**
   A natural number \( n > 1 \) is called "heavy" if it is coprime with the sum of its divisors, denoted by \( \sigma(n) \). This means \( \gcd(n, \sigma(n)) = 1 \).

2. **Analyzing powers of 2:**
   For any natural number \( n \), if \( n = 2^k \) (a power of 2), then \( \sigma(2^k) = 2^{k+1} - 1 \). Since \( 2^{k+1} - 1 \) is odd, \( \gcd(2^k, 2^{k+1} - 1) = 1 \). Therefore, all powers of 2 are heavy.

3. **Analyzing other even numbers:**
   For any even number \( n \) that is not a power of 2, \( n \) has at least one odd divisor. The sum of the divisors \( \sigma(n) \) will be even because it includes the even number \( n \) itself. Hence, \( \gcd(n, \sigma(n)) \neq 1 \), and such \( n \) cannot be heavy.

4. **Analyzing odd numbers:**
   For odd numbers \( n \), \( \sigma(n) \) is also odd because the sum of odd numbers is odd. Therefore, \( \gcd(n, \sigma(n)) = 1 \), making all odd numbers heavy.

5. **Finding the maximal number of consecutive heavy numbers:**
   - We need to find the longest sequence of consecutive numbers where each number is heavy.
   - Since all powers of 2 are heavy and all odd numbers are heavy, we need to consider sequences involving these numbers.
   - The only powers of 2 that are consecutive are \( 2 \) and \( 4 \). Therefore, we can consider the sequence \( 2, 3, 4, 5 \):
     - \( 2 \) is heavy (as shown above).
     - \( 3 \) is heavy (odd number).
     - \( 4 \) is heavy (as shown above).
     - \( 5 \) is heavy (odd number).

6. **Conclusion:**
   The sequence \( 2, 3, 4, 5 \) is a maximal sequence of consecutive heavy numbers, and its length is 4.

The final answer is \(\boxed{4}\).