18. There are 2012 students standing in a row, numbered from left to right as $1, 2, \cdots \cdots 2012$. In the first round, they report numbers from left to right as “1, 2”, and those who report 2 stay; from the second round onwards, each time the remaining students report numbers from left to right as “1, 2, 3”, and those who report 3 stay, until only one student remains. What is the number of the last remaining student?
Analysis: The key point is the problem of regularity and intelligent and interesting questions.
The first time, students report numbers "1,2" from left to right, and those who report 2 stay, so the ones left after the first round must be the people at positions that are multiples of 2, which are $2,4,6, \cdots \cdots 2012$.

The second time, the students who stayed report numbers "1,2,3" from left to right, and those who report 3 stay, so the ones left must be the people at positions that are multiples of 3. By finding the pattern, it is not difficult to see that the ones left in the next round must be at positions that are multiples of $3 \times 3$, and the ones left in the following round must be at positions that are multiples of $3 \times 3 \times 3$.....

Thus, the 1006 people who stayed after the first round are re-arranged in order, forming a queue from $1-1006$, so the one left at the end must be the $3 \times 3 \times 3 \times 3 \times 3 \times 3=729$th person. Therefore, in the entire queue, this person is at the $729 \times 2=1458$th position.