Find all positive integers $n$ for which both $837 + n$ and $837 - n$ are cubes of positive integers.
To find all positive integers \( n \) for which both \( 837 + n \) and \( 837 - n \) are cubes of positive integers, we can proceed as follows:

1. Let \( 837 + n = y^3 \) and \( 837 - n = x^3 \) for some positive integers \( x \) and \( y \). This implies:
   \[
   y^3 - x^3 = (837 + n) - (837 - n) = 2n
   \]
   Therefore, we have:
   \[
   y^3 - x^3 = 2n
   \]

2. Using the factorization of the difference of cubes, we get:
   \[
   y^3 - x^3 = (y - x)(y^2 + yx + x^2)
   \]
   Thus, we can write:
   \[
   2n = (y - x)(y^2 + yx + x^2)
   \]

3. Since \( n \) is a positive integer, \( 2n \) must be an even integer. Therefore, \( (y - x)(y^2 + yx + x^2) \) must be even. 

4. We know that \( 837 \pm n \) are both cubes, so we need to find \( x \) and \( y \) such that both \( 837 - n \) and \( 837 + n \) are cubes. We can express \( n \) in terms of \( x \) and \( y \):
   \[
   n = \frac{y^3 - x^3}{2}
   \]

5. We need to find pairs \((x, y)\) such that \( 837 - x^3 \) and \( 837 + y^3 \) are both positive integers. This means \( x^3 < 837 \) and \( y^3 > 837 \).

6. We need to factorize \( 2n = (y - x)(y^2 + yx + x^2) \) and check for possible values of \( x \) and \( y \). The prime factorization of 837 is:
   \[
   837 = 3^3 \times 31
   \]
   Therefore, \( 2n = 2 \times 837 = 2 \times 3^3 \times 31 = 2 \times 27 \times 31 = 2 \times 837 = 1674 \).

7. We need to find pairs \((x, y)\) such that:
   \[
   (y - x)(y^2 + yx + x^2) = 1674
   \]

8. By checking possible pairs, we find that:
   \[
   (x, y) = (7, 11)
   \]
   This gives:
   \[
   837 - n = 7^3 = 343 \quad \text{and} \quad 837 + n = 11^3 = 1331
   \]
   Solving for \( n \):
   \[
   837 - n = 343 \implies n = 837 - 343 = 494
   \]

Therefore, the only positive integer \( n \) that satisfies the conditions is \( n = 494 \).

The final answer is \( \boxed{494} \).