Example 3. Arrange $1,2,3, \cdots, 1989$ in a circle, and starting from 1, cross out every other number. (That is, keep 1, cross out 2, keep 3, cross out $4, \cdots$), and repeat this process multiple times until only one number remains. What is this number?
Solving: Arranging so many numbers in a circle and performing deletions makes it difficult to see the pattern. Therefore, one should start with simpler cases. Suppose $n$ numbers are arranged in a circle, and the specified operations are performed. Let the last remaining number be denoted as $T_{n}$. The following table lists the initial cases:
Carefully observing, it is not difficult to find that the values of $T$ do not include even numbers (in fact, all even numbers are eliminated in the first round). When $n=1,2,4,8,16$, the values of $T$ are all 1. It is also not difficult to notice an interesting phenomenon: when $2^{k}<n<2^{k+1}$, the values of $T_{n}$ form an arithmetic sequence of odd numbers: 3, 5, 7, ... Thus, we can reasonably conjecture that when $m$ is a natural number less than $2^{k-1}$, $T_{n}=2m+1$.
(Conjectures (I) and (II) are not difficult to prove; please verify them yourself.)
When $n=1989$, since $1989=2^{10}+965$, we have $m=965$, thus $T_{n}=2m+1=1931$. Therefore, the last remaining number is 1931.

From these two examples, we can see that for complex problems, one can start with simpler cases to identify patterns, providing clues and a good start for solving more complex original problems.