Determine all natural numbers $n$ such that $9^n - 7$ can be represented as a product of at least two consecutive natural numbers.
To determine all natural numbers \( n \) such that \( 9^n - 7 \) can be represented as a product of at least two consecutive natural numbers, we need to find \( n \) such that \( 9^n - 7 = k(k+1) \) for some natural number \( k \).

1. **Rewrite the equation:**
   \[
   9^n - 7 = k(k+1)
   \]
   This implies:
   \[
   9^n - 7 = k^2 + k
   \]

2. **Consider the quadratic equation:**
   \[
   k^2 + k - (9^n - 7) = 0
   \]
   For \( k \) to be an integer, the discriminant of this quadratic equation must be a perfect square. The discriminant \(\Delta\) is given by:
   \[
   \Delta = 1 + 4(9^n - 7)
   \]
   \[
   \Delta = 4 \cdot 9^n - 27
   \]

3. **Check for perfect square:**
   We need \( 4 \cdot 9^n - 27 \) to be a perfect square. Let:
   \[
   4 \cdot 9^n - 27 = m^2
   \]
   for some integer \( m \). Rearrange to get:
   \[
   4 \cdot 9^n - m^2 = 27
   \]
   \[
   4 \cdot 9^n - 27 = m^2
   \]
   \[
   4 \cdot 9^n = m^2 + 27
   \]

4. **Check small values of \( n \):**
   - For \( n = 1 \):
     \[
     4 \cdot 9^1 = 36
     \]
     \[
     36 - 27 = 9
     \]
     \[
     m^2 = 9 \implies m = 3
     \]
     \[
     9^1 - 7 = 2 \quad \text{(not a product of two consecutive numbers)}
     \]

   - For \( n = 2 \):
     \[
     4 \cdot 9^2 = 324
     \]
     \[
     324 - 27 = 297
     \]
     \[
     m^2 = 297 \quad \text{(not a perfect square)}
     \]

   - For \( n = 3 \):
     \[
     4 \cdot 9^3 = 2916
     \]
     \[
     2916 - 27 = 2889
     \]
     \[
     m^2 = 2889 \quad \text{(not a perfect square)}
     \]

5. **General case for \( n > 3 \):**
   If \( n > 3 \), then \( 4 \cdot 9^n - 27 \) grows very large and it is unlikely to be a perfect square. We can use modular arithmetic to show that \( 4 \cdot 9^n - 27 \) modulo 81 does not yield a perfect square.

   - For \( n = 4 \):
     \[
     4 \cdot 9^4 = 26244
     \]
     \[
     26244 - 27 = 26217
     \]
     \[
     m^2 = 26217 \quad \text{(not a perfect square)}
     \]

6. **Conclusion:**
   After checking small values and considering the general case, we find that there are no natural numbers \( n \) such that \( 9^n - 7 \) can be represented as a product of at least two consecutive natural numbers.

The final answer is \( \boxed{\text{None}} \)