9. (6 points) Three mice found a pile of peanuts and agreed to come and share them equally the next day. The next day, the first mouse arrived the earliest. He found that the peanuts could not be divided equally, so he ate one, and the remaining could be divided into 3 equal parts, and he took one part. The second and third mice came one after another, encountering the same problem, and took the same method, each eating one and then dividing the remaining peanuts into three parts, taking one part. So, what is the minimum number of peanuts in this pile?
【Solution】Solution:
(1) The last rat takes 1 grain,
Before the last rat takes: $1 \times 3+1=4$ (grains);
Before the second rat takes: $4 \times 3 \div 2+1=7$ (grains);
Before the first rat takes: $7 \times 3 \div 2+1=12.5$ (grains) cannot be divided evenly, discard.
(2) The last rat takes 2 grains,
Before the last rat takes: $2 \times 3+1=7$ (grains);
Before the second rat takes: $7 \times 3 \div 2+1=12.5$ cannot be divided evenly, discard.
(3) The last rat takes 3 grains,

Before the last rat takes: $3 \times 3+1=10$ (grains);
Before the second rat takes: $10 \times 3 \div 2+1=16$ (grains);
Before the first rat takes: $16 \times 3 \div 2+1=25$ (grains), which meets the requirement.

Therefore, the initial pile of peanuts has at least 25 grains.
Answer: The pile of peanuts has at least 25 grains.