9. (12 points) Four cats, four dogs, and four mice are kept in 12 cages. If a cat and a mouse are in the same column, the cat will meow non-stop; if a mouse is flanked by two cats on either side, the mouse will squeak non-stop; if a dog is flanked by a cat and a mouse on either side, the dog will bark non-stop. In all other cases, the animals remain silent. One day, the cages numbered 3, 4, 6, 7, 8, and 9 are very noisy, while the other cages are very quiet. What is the sum of the cage numbers where the four dogs are located? $\qquad$
【Analysis】According to the problem, we get that $1$ and $12$ are mice, and $6$ and $7$ are cats. We can use the assumption method to reason from the 8 possible scenarios. After determining 8, we can further determine 2, and then infer 3 from 2. After 3 is determined, we can further determine 4 and 10, and then solve the problem.

【Solution】According to the analysis, since $6$ and $7$ are very noisy, we get that $1$ and $12$ are mice, and $6$ and $7$ are cats. If 8 is a dog, then 9 must be a mouse, but 9 as a mouse cannot be sandwiched by cats, which is not valid; if 8 is a mouse, then 9 is a cat, and since 9 is also noisy, we get that 3 is a mouse. Since 3 is noisy, we get that 2 and 4 are both cats, which means there are already 5 cats, which is not valid;

Therefore, 8 can only be a cat. Since 8 is noisy, we get that 2 is a mouse. If 3 is a cat, then 9 is a mouse, and 10 must also be a cat, which is not valid;

If 3 is a mouse, it cannot be noisy, which is not valid;
So 3 is a dog, then 4 is a cat. Since 4 is noisy, we get that 10 is a mouse. Therefore, the rest of the unknowns are all dogs. In summary, the sum of the numbers of the dogs is: $3+5+9+11=28$.

Therefore, the answer is: 28.