We have $a_{n+1}a_{n-2}= 1989+a_n a_{n-1}$. Iterating this, we have $a_n a_{n-3} = 1989+a_{n-1}a_{n-2}$. Subtracting, and doing some algebra, we arrive at \[ a_{n-2}\left(a_{n+1}+a_{n-1}\right) = a_n\left(a_{n-1}+a_{n-3}\right)\Leftrightarrow \frac{a_{n+1}+a_{n-1}}{a_{n-1}+a_{n-3}} = \frac{a_n}{a_{n-2}}. \]We now telescope this identity, based on the parity of $n$. If $n$ is odd, we iterate, and obtain \[ \frac{a_{n+1}+a_{n-1}}{a_4+a_2} = \frac{a_n}{a_3} \implies a_{n+1} = 11a_n -a_{n-1}. \]Moreover, if $n$ is even, the same reasoning now yields \[ a_{n+1} = 200a_n -a_{n-1}. \]From these recursions; it is evident $\left(a_n\right)_{n\ge 1}\subset \mathbb{N}$, as requested.