 Is it true that there are $130$ consecutive natural numbers, such that each of them has exactly $900$ natural divisors?
1. **Understanding the problem**: We need to determine if there exist 130 consecutive natural numbers such that each of them has exactly 900 natural divisors.

2. **Analyzing the number of divisors**: The number of divisors of a number \( n \) can be determined from its prime factorization. If \( n \) has the prime factorization \( n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k} \), then the number of divisors \( d(n) \) is given by:
   \[
   d(n) = (e_1 + 1)(e_2 + 1) \cdots (e_k + 1)
   \]
   We need \( d(n) = 900 \).

3. **Factorizing 900**: We factorize 900 to understand the possible forms of \( n \):
   \[
   900 = 2^2 \cdot 3^2 \cdot 5^2
   \]
   The number 900 can be expressed as a product of integers in several ways, such as:
   \[
   900 = 9 \cdot 10 \cdot 10 = 3^2 \cdot 10 \cdot 10
   \]
   This means the exponents in the prime factorization of \( n \) must satisfy:
   \[
   (e_1 + 1)(e_2 + 1) \cdots (e_k + 1) = 900
   \]

4. **Considering the modulo condition**: We need to check if there can be 130 consecutive numbers each having exactly 900 divisors. Note that for any number \( n \), if \( n \equiv 64 \pmod{128} \), then \( \nu_2(n) = 6 \) (where \( \nu_2(n) \) denotes the highest power of 2 dividing \( n \)).

5. **Divisors of \( n \equiv 64 \pmod{128} \)**: If \( n \equiv 64 \pmod{128} \), then \( n = 2^6 \cdot m \) where \( m \) is odd. The number of divisors of \( n \) is:
   \[
   d(n) = (6+1) \cdot d(m) = 7 \cdot d(m)
   \]
   Since 7 is a prime number, \( d(n) \) must be a multiple of 7. However, 900 is not a multiple of 7 (since \( 900 \div 7 \) is not an integer).

6. **Conclusion**: Since one of the 130 consecutive numbers must be \( n \equiv 64 \pmod{128} \), and such a number cannot have exactly 900 divisors, it is impossible for all 130 consecutive numbers to each have exactly 900 divisors.

\[
\boxed{\text{no}}
\]