Solution:

Let $a_n$ be the first number remaining after $n$ iterations, so $a_0 = 1$, $a_1 = 2$, $a_2 = 3$, $a_3 = 5$, etc. We claim that:

$\displaystyle a_{n+1} = \frac{3}{2} a_n$ if $a_n$ is even, and
$\displaystyle a_{n+1} = \frac{3}{2}(a_n + 1) - 1$ if $a_n$ is odd.

We use induction on $n$.

Suppose $a_n = 2N$. Consider the number $3N$. There are initially $N$ smaller numbers $\equiv 1 \pmod{3}$. So after the first iteration, it will lie in the $2N$th place. Hence, it will lie in first place after $n+1$ iterations.

Similarly, suppose $a_n = 2N + 1$. Consider $3N + 2$. There are initially $N + 1$ smaller numbers $\equiv 1 \pmod{3}$. So after the first iteration, it will lie in the $2N + 1$st place. Hence, it will lie in first place after $n+1$ iterations. That completes the induction.

We may now calculate successively the members of the sequence: $1, 2, 3, 5, 8, 12, 18, 27, 41, 62, 93, 140, 210, 315, 473, 710, 1065, 1598, 2397$. Hence $1598$ is the last surviving number from $1, 2, \ldots, 2002$.