7. There are less than 100 cells in a row, and at the beginning, only the cells at both ends contain a chess piece each. Several students take turns placing chess pieces in the cells. Each person places one piece at a time, and it must be placed in the middle of two adjacent cells that already contain pieces (for example, if the 3rd and 7th cells from left to right contain pieces, and the 4th, 5th, and 6th cells do not, a piece can be placed in the 5th cell; but if the 4th and 7th cells contain pieces, and the 5th and 6th cells do not, neither the 5th nor the 6th cell can be used). Each of these students placed 9 pieces, resulting in each cell containing exactly one piece. How many students are there? $\qquad$
【Solution】Solution: According to the problem, if there are an even number of empty spaces between two adjacent chess pieces, no more pieces can be placed in between. Therefore, to place a chess piece in every space, the number of empty spaces between each pair of adjacent pieces should be odd after each placement.

In the first round, only 1 piece can be placed in the middle, dividing the spaces into the front half and the back half. In the second round, 1 piece can be placed in the middle of each half, totaling 2 pieces, which then divides the spaces into 4 parts. In the third round, 1 piece can be placed in the middle of each part, totaling 4 pieces, and so on. The total number of pieces placed should be the sum of the geometric sequence $1,2,4,8, \cdots$. Since each person places 9 times, this sum is a multiple of 9 and cannot exceed 100. By enumeration, we get $1+2+4+8+16+32=63$, which meets the condition, so there are $63 \div 9=7$ students, and the distribution of the chess pieces is as follows:
1,65
$1,33,65$
$1,17,33,49,65$
$1,9,17,25,33,41,49,57,65$,
$\cdots$
Therefore, the answer is 7.