Three, (25 points) Divide $1,2, \cdots, 9$ into three groups, each containing three numbers, such that the sum of the numbers in each group is a prime number.
(1) Prove that there must be two groups with equal sums;
(2) Find the number of all different ways to divide them.


Three, (1) Since the sum of three different numbers in $1,2, \cdots, 9$ is between 6 and 24, the prime numbers among them are only 7, 11, 13, 17, 19, 23, these six. Now, these six numbers are divided into two categories based on their remainders when divided by 3:
$A=\{7,13,19\}$, where each number leaves a remainder of 1 when divided by 3;
$B=\{11,17,23\}$, where each number leaves a remainder of 2 when divided by 3.
Suppose the sums $p_{a}, p_{b}, p_{c}$ of the three groups $A, B, C$ are distinct primes.

Since $p_{a}+p_{b}+p_{c}=1+2+\cdots+9=45$ is divisible by 3, the three sums $p_{a}, p_{b}, p_{c}$ must be of the same type. Note that the sum of the three numbers in class $A$ is $7+13+19=39 \neq 45$, which is a contradiction.
Therefore, two of the three sums must be equal.
(2) According to (1), expressing 45 as the sum of three numbers from 7, 11, 13, 17, 19, 23 (with two of the numbers being equal), there are only four cases:
(1) $19+19+7$, (2) $17+17+11$,
(3) $13+13+19$, (4) $11+11+23$.
Since there are 5 odd numbers in $1,2, \cdots, 9$, one of the three groups must consist of three odd numbers, and the other two groups must each have one odd number.

For case (1), the group with a sum of 7 can only be $\{1,2,4\}$. Dividing the remaining six numbers $3,5,6,7,8,9$ into two groups with a sum of 19, and one of these groups must consist of all odd numbers, there is only one unique division: $\{3,7,9\}$ and $\{5,6,8\}$.

For case (2), if the group of three odd numbers is $\{1,7,9\}$, then the other two groups can be $\{4,5,8\},\{2,3,6\}$ or $\{3,6,8\},\{2,4,5\}$;

If the group of three odd numbers is $\{3,5,9\}$, then the other two groups can be $\{2,8,7\},\{1,4,6\}$ or $\{4,6,7\},\{1,2,8\}$;

If the group of three odd numbers is $\{1,3,7\}$, then the other two groups can be $\{2,6,9\},\{4,5,8\}$.
This results in 5 different divisions.
For case (3), if the group of three odd numbers is $\{3,7,9\}$, then the other two groups can be $\{1,4,8\},\{2,5,6\}$;

If the group of three odd numbers is $\{1,3,9\}$, then the other two groups can be $\{2,4,7\},\{5,6,8\}$ or $\{2,5,6\},\{4,7,8\}$;

If the group of three odd numbers is $\{1,5,7\}$, then the other two groups can be $\{3,4,6\},\{2,8,9\}$ or $\{2,3,8\},\{4,6,9\}$.
This results in 5 different divisions.
For case (4), the group with a sum of 23 can only be $\{6,8,9\}$, then the other two groups can be $\{1,3,7\},\{2,4,5\}$.
In total, there are $1+5+5+1=12$ different divisions.
(Tao Pingsheng, Department of Mathematics and Computer Science, Jiangxi Science and Technology Normal University, 330013)