Let $x, y$ be the two numbers. So $x + y =n$. We have $x | x+y+6$ which implies $x|y+6$. The same way we get $y|x+6$. Suppose, WLOG, that $y \geq x$. The condition $y | x+6$ implies that $y \leq x+6$, or $y - x \leq 6$. The rest is case work: $y -x = 6$. Then $x|y+6 = x+12$, and so $x|12$, giving $n \in \{ 8,10,12,14,18,30\}$. $y-x = 5$. Then $x|11$. No solutions for $n$. $y - x = 4$. Then $x|10$, giving $n = 12$ as the only solution. $y -x = 3$. Then $x|9$. No solutions for $n$. $y - x = 2$. Then $x | 8$, , giving $n = 6$ as the only solution. $y -x = 1$. Then $x|7$. No solutions for $n$. $y -x = 0$. Then $x|6$, giving $n \in \{ 2, 4, 6, 12 \}$. And finally, $\{2,4,6,8,10,12,14,18,30\}$ are the only possible values for $n$.