\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