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 $\boxed{\text{None}}$