$2 \cdot 80$ integer sequence $a_{1}, a_{2}, a_{3}, \cdots$ is defined as follows: $a_{1}=1$, for $n \geqslant 1, a_{n+1}$ is the smallest integer greater than $a_{n}$ such that for any $i, j, k$ (not necessarily distinct) in $\{1,2, \cdots, n+1\}$, $a_{i} + a_{j} \neq 3 a_{k}$, determine $a_{1998}$.
（39th International Mathematical Olympiad Preliminary Question, 1998）
[Solution] Given $a_{1}=1$, then $a_{2}=3, a_{3}=4, a_{4}=7$, which satisfies the problem's requirements. This is because the sum of any two numbers (which can be the same) from $1,3,4,7$ is $1,2,4,5,6,7,8,10,11,14$, and their remainders when divided by 9 are $1,2,4,5,6,7,8$, while the remainders of $3 a_{k}$ when divided by 9 are 0, 3.

Let the first $4 n$ terms of the sequence $\left\{a_{m}\right\}$ in $1,2, \cdots, 9 n$ be those numbers whose remainders when divided by 9 are $1,3,4,7$. Then, after adding $9 n+1$ to the first $4 n$ terms of the sequence $\left\{a_{m}\right\}$, we have
$$
\begin{array}{c}
a_{i}+a_{j} \equiv 1,2,4,5,6,7,8 \quad(\bmod 9), \\
3 a_{k} \equiv 0,9 \quad(\bmod 9),
\end{array}
$$

Thus,
$$
a_{4 n+1}=9 n+1 \text {. }
$$

Since
$$
(9 n+2)+1=3(3 n+1),
$$

and $3 n+1 \equiv 1,4,7 \quad(\bmod 9), 1 \equiv 1 \quad(\bmod 9)$,
therefore
$$
9 n+2 \notin\left\{a_{m}\right\} \text {. }
$$

Similarly, it can be shown that $9 n+3,9 n+4,9 n+7 \in\left\{a_{m}\right\}$,
and because $(9 n+5)+7=3(3 n+4)$,
i.e., $(9 n+8)+4=3(3 n+4)$,
or $(9 n+6)+(9 n+6)=3(6 n+4)$,
or $(9 n+9)+3=3(3 n+4)$,
since $3 n+4 \equiv 1,4,7(\bmod 9)$,
and $6 n+4 \equiv 1,4,7(\bmod 9)$,
so $3 n+4,6 n+4 \in\left\{a_{n}\right\}$,
thus $9 n+5,9 n+6,9 n+8,9 n+9 \notin\left\{a_{n}\right\}$.
Therefore, $\left\{a_{m}\right\}$ consists of the natural numbers that leave remainders of $1,3,4,7$ when divided by 9.
Since $1998=499 \times 4+2$, we have $a_{1998}=499 \times 9+3=4494$.