We claim that the sequence consists of all positive integers that are $1,3,4,7\pmod 9$. By considering $\pmod 9$ this sequence works. This would give $\boxed{a_{1998} = 4494}$. We prove by induction that this is the only solution.
First, notice that $a_1 = 1, a_2 = 3, a_3 = 4, a_4 = 7, a_5 = 10, a_6 = 12,$.
Suppose that $a_1, a_2, \ldots, a_{n-1}$ consists of positive integers only $1,3,4,7\pmod 9$ (so in fact, all positive integers $1,3,4,7\pmod 9$ at most $a_{n-1}$). We will show that $a_n\equiv \{1,3,4,7\}\pmod 9$.
If $a_n = 9k$, then notice that $9k + (9(k-1) + 3) = 3(6k-2)$, contradiction.
If $a_n = 9k +2$, then we have $(9k+2) + (9k+1) = 3(6k+1)$, contradiction.
If $a_n = 9k + 5$, then we have $(9k+5) + (9k-2) = 3(6k+1)$, contradiction.
If $a_n = 9k + 6$, then $(9k+6) + (9k+6) = 3(6k + 4)$, contradiction.
If $a_n = 9k+8$, then $(9k+8) + (9k-5) = 3(6k +1)$, contradiction.
Thus, $a_n$ cannot be $0,2,5,6,$ or $8$ mod $9$, which proves $a_n \in \{1,3,4,7\}\pmod 9$, so the induction is complete.
We now conclude $a_i\in \{1,3,4,7\}\pmod 9$, which finishes.