Find all natural numbers $n$ for which $n + 195$ and $n - 274$ are  perfect cubes.
1. Let \( n + 195 = x^3 \) and \( n - 274 = y^3 \). This implies:
   \[
   n = x^3 - 195 \quad \text{and} \quad n = y^3 + 274
   \]
   Equating the two expressions for \( n \), we get:
   \[
   x^3 - 195 = y^3 + 274
   \]
   Simplifying, we obtain:
   \[
   x^3 - y^3 = 469
   \]

2. We can factorize \( x^3 - y^3 \) using the difference of cubes formula:
   \[
   x^3 - y^3 = (x - y)(x^2 + xy + y^2)
   \]
   Therefore, we have:
   \[
   (x - y)(x^2 + xy + y^2) = 469
   \]

3. Since 469 is a product of primes \( 7 \times 67 \), we consider the possible factor pairs:
   - \( x - y = 1 \) and \( x^2 + xy + y^2 = 469 \)
   - \( x - y = 7 \) and \( x^2 + xy + y^2 = 67 \)

4. **Case 1: \( x - y = 1 \)**
   \[
   x - y = 1 \implies x = y + 1
   \]
   Substituting \( x = y + 1 \) into \( x^2 + xy + y^2 = 469 \):
   \[
   (y + 1)^2 + (y + 1)y + y^2 = 469
   \]
   Simplifying:
   \[
   y^2 + 2y + 1 + y^2 + y + y^2 = 469
   \]
   \[
   3y^2 + 3y + 1 = 469
   \]
   \[
   3y^2 + 3y - 468 = 0
   \]
   Dividing by 3:
   \[
   y^2 + y - 156 = 0
   \]
   Solving this quadratic equation using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   y = \frac{-1 \pm \sqrt{1 + 624}}{2}
   \]
   \[
   y = \frac{-1 \pm \sqrt{625}}{2}
   \]
   \[
   y = \frac{-1 \pm 25}{2}
   \]
   This gives us two solutions:
   \[
   y = \frac{24}{2} = 12 \quad \text{and} \quad y = \frac{-26}{2} = -13
   \]
   Since \( y \) must be a natural number, we take \( y = 12 \). Then:
   \[
   x = y + 1 = 12 + 1 = 13
   \]
   Substituting \( x \) and \( y \) back into the expression for \( n \):
   \[
   n = y^3 + 274 = 12^3 + 274 = 1728 + 274 = 2002
   \]

5. **Case 2: \( x - y = 7 \)**
   \[
   x - y = 7 \implies x = y + 7
   \]
   Substituting \( x = y + 7 \) into \( x^2 + xy + y^2 = 67 \):
   \[
   (y + 7)^2 + (y + 7)y + y^2 = 67
   \]
   Simplifying:
   \[
   y^2 + 14y + 49 + y^2 + 7y + y^2 = 67
   \]
   \[
   3y^2 + 21y + 49 = 67
   \]
   \[
   3y^2 + 21y - 18 = 0
   \]
   Dividing by 3:
   \[
   y^2 + 7y - 6 = 0
   \]
   Solving this quadratic equation using the quadratic formula:
   \[
   y = \frac{-7 \pm \sqrt{49 + 24}}{2}
   \]
   \[
   y = \frac{-7 \pm \sqrt{73}}{2}
   \]
   Since \( \sqrt{73} \) is not an integer, there are no natural number solutions for \( y \) in this case.

Thus, the only solution is \( n = 2002 \).

The final answer is \( \boxed{2002} \)