7. (10 points) There are 11, 12, and 17 balls of red, yellow, and blue colors respectively. Each operation can replace 2 balls of different colors with 2 balls of the third color. During the operation process, the maximum number of red balls can be $\qquad$.
【Answer】Solution: The remainders when the number of the three types of balls is divided by 3 are $2.0$, $2$, respectively. After any operation, the remainders when divided by 3 all increase by 2. Therefore, the remainders when the number of yellow balls and blue balls is divided by 3 cannot be the same, meaning it is impossible to have 0 yellow balls and 0 blue balls at the same time, so the number of red balls cannot be 40.

Upon verification, the first two operations change the red balls and blue balls into yellow balls, resulting in the ball counts being $9$, $16$, $15$; then changing the yellow balls and blue balls into red balls results in the ball counts being $39$, $1$, $0$. Therefore, during the operations, the maximum number of red balls is 39.

Answer: The maximum number of red balls is 39.
So the answer is: 39.