We need to find all positive integers $n$ for which there exist positive integers $x, y$ so that $837 + n = x^3$ and $837 - n = y^3$. Adding these equations gives
$$1674 = x^3 + y^3 = (x + y)(x^2 - x y + y^2).$$
Let $u = x + y$ and $v = x^2 - x y + y^2 = (x + y)^2 - 3 x y = u^2 - 3 x y$, then $3 x y = u^2 - v$ and we see that $u$ is divisible by $3$ if and only if $v$ is divisible by $3$.

As $1674 = 2 \cdot 3^3 \cdot 31 = u \cdot v$, there exist positive integers $u_1$ and $v_1$ such that $u = 3 u_1$ and $v = 3 v_1$. The above equations translate into
$$186 = u_1 v_1, \quad x + y = 3 u_1, \quad x y = 3 u_1^2 - v_1.$$
As $x, y > 0$ we get $3 u_1^2 > v_1$, hence $3 u_1^3 > u_1 v_1 = 186$ and so $u_1 \ge 4$. On the other hand, the AM-GM inequality gives $(x + y)^2 \ge 4 x y$, which translates into $9 u_1^2 \ge 12 u_1^2 - 4 v_1$, i.e. $4 v_1 \ge 3 u_1^2 > 4 u_1$, the last inequality because $3 u_1 > 4$.

The factors of $186$ are $1, 2, 3, 6, 31, 62, 93, 186$. Using $v_1 > u_1 \ge 4$ we see that we must have $u_1 = 6$ and $v_1 = 31$. This leads to $x + y = 18$ and $x y = 77$ from which we obtain the quadratic equation $x^2 - 18 x + 77 = 0$. The two solutions are $11$ and $7$. As $n > 0$ we have $x > y$ and so $(x, y) = (11, 7)$. Therefore, $n = 837 - 7^3 = 11^3 - 837 = 494$.

We need to find all positive integers $n$ for which there exist positive integers $x, y$ so that $837 + n = x^3$ and $837 - n = y^3$. From $n > 0$ and $y > 0$ we obtain $1 \le n \le 836$. This implies that $838 \le x^3 = 837 + n \le 1673$. Because $9^3 = 729 < 838$ and $1673 < 1728 = 12^3$ we can only have $x = 10$ or $x = 11$. If $x = 10$, we obtain $n = x^3 - 837 = 163$ which implies $y^3 = 837 - n = 674$, but this is not the cube of an integer. With $x = 11$ we find $n = x^3 - 837 = 1331 - 837 = 494$ and $y^3 = 837 - n = 837 - 494 = 343 = 7^3$, hence $y = 7$. Therefore, $(x, y) = (11, 7)$ is the only solution and $n = 494$.