\begin{align*} x^3 + y^3 &= (x - y)(x^2 + xy + y^2) \\ &= (x - y)((x - y)^2 + 3xy ) \\ &= 999 \end{align*}As $x$ and $y$ are positive integers, we can easily factor 999 and try all the possible combinations. Also note that $(x - y) < ((x - y)^2 + 3xy )$ \begin{align*} (x - y)((x - y)^2 + 3xy ) &= 999 \\ &= 1 \cdot 999 \\ &= 3 \cdot 333 \\ &= 9 \cdot 111 \\ &= 27 \cdot 37 \end{align*}Case 1: $ x - y = 1 $ $ ((x - y)^2 + 3xy ) = 999 $ Solving gives no solution in $N$ Case 2: $ x - y = 3 $ $ ((x - y)^2 + 3xy ) = 333 $ This gives one solution (x = 12, y = 9) Case 3: $ x - y = 9 $ $ ((x - y)^2 + 3xy ) = 111 $ This gives one solution (x = 10, y = 1) Case 4: $ x - y =27 $ $ ((x - y)^2 + 3xy ) = 37 $ Solving gives no solution in $N$ Therefore only (10, 1) and (12, 9) are solutions