Find $n$ such that $n - 76$ and $n + 76$ are both cubes of positive integers.
1. We need to find an integer \( n \) such that both \( n - 76 \) and \( n + 76 \) are cubes of positive integers.
2. Let \( n - 76 = a^3 \) and \( n + 76 = b^3 \) for some positive integers \( a \) and \( b \).
3. Subtracting the first equation from the second, we get:
   \[
   (n + 76) - (n - 76) = b^3 - a^3
   \]
   Simplifying, we have:
   \[
   152 = b^3 - a^3
   \]
4. We can factor the difference of cubes:
   \[
   b^3 - a^3 = (b - a)(b^2 + ba + a^2)
   \]
   Therefore:
   \[
   152 = (b - a)(b^2 + ba + a^2)
   \]
5. We need to find pairs \((a, b)\) such that the product of \((b - a)\) and \((b^2 + ba + a^2)\) equals 152. We can try different values for \( b - a \) and solve for \( a \) and \( b \).

6. Trying \( b - a = 1 \):
   \[
   152 = 1 \cdot (b^2 + ba + a^2)
   \]
   \[
   b^2 + ba + a^2 = 152
   \]
   This does not yield integer solutions for \( a \) and \( b \).

7. Trying \( b - a = 2 \):
   \[
   152 = 2 \cdot (b^2 + ba + a^2)
   \]
   \[
   b^2 + ba + a^2 = 76
   \]
   This does not yield integer solutions for \( a \) and \( b \).

8. Trying \( b - a = 4 \):
   \[
   152 = 4 \cdot (b^2 + ba + a^2)
   \]
   \[
   b^2 + ba + a^2 = 38
   \]
   This does not yield integer solutions for \( a \) and \( b \).

9. Trying \( b - a = 8 \):
   \[
   152 = 8 \cdot (b^2 + ba + a^2)
   \]
   \[
   b^2 + ba + a^2 = 19
   \]
   This does not yield integer solutions for \( a \) and \( b \).

10. Trying \( b - a = 19 \):
    \[
    152 = 19 \cdot (b^2 + ba + a^2)
    \]
    \[
    b^2 + ba + a^2 = 8
    \]
    This does not yield integer solutions for \( a \) and \( b \).

11. Trying \( b - a = 38 \):
    \[
    152 = 38 \cdot (b^2 + ba + a^2)
    \]
    \[
    b^2 + ba + a^2 = 4
    \]
    This does not yield integer solutions for \( a \) and \( b \).

12. Trying \( b - a = 76 \):
    \[
    152 = 76 \cdot (b^2 + ba + a^2)
    \]
    \[
    b^2 + ba + a^2 = 2
    \]
    This does not yield integer solutions for \( a \) and \( b \).

13. Trying \( b - a = 152 \):
    \[
    152 = 152 \cdot (b^2 + ba + a^2)
    \]
    \[
    b^2 + ba + a^2 = 1
    \]
    This does not yield integer solutions for \( a \) and \( b \).

14. After trying all possible values, we find that \( a = 4 \) and \( b = 6 \) work:
    \[
    b - a = 2
    \]
    \[
    b^2 + ba + a^2 = 76
    \]
    \[
    6^2 + 6 \cdot 4 + 4^2 = 36 + 24 + 16 = 76
    \]
    Therefore, \( n = a^3 + 76 = 4^3 + 76 = 64 + 76 = 140 \).

The final answer is \( \boxed{140} \).