Example 2.2.3 In the sequence $1,9,81, \cdots, 9^{2005}$, delete the terms whose highest digit is 9. How many terms are left in the remaining sequence?
Solution: If the first digit of a term in the sequence is 9, then the first digit of the previous term is 1, and the number of digits in the two adjacent terms is the same. Conversely, this is also true, meaning there is a one-to-one correspondence between the terms with the first digit 9 and the two adjacent terms with the same number of digits. By removing the latter of the two adjacent terms with the same number of digits, the remaining sequence has numbers with exactly $1, 2, \cdots, t$ digits, totaling $t$ terms, where $t$ is the number of digits in $9^{2005}$, so $t=\left[\lg 9^{2005}\right]+1=[4010 \times \lg 3]+1=[4010 \times 0.47712]+1=1914$.