Three piles of stones have the numbers 21, 10, and 11, respectively. Now, the following operation is performed: each time, 1 stone is taken from any two piles, and then these 2 stones are added to the other pile. The question is: 
(1) Can the number of stones in the three piles be 4, 14, and 24 after several such operations?
(2) Can the number of stones in the three piles all be 14 after several such operations?
If the requirements can be met, please complete it with the minimum number of operations; if they cannot be met, please explain the reason.
(1) It can be achieved.
The minimum number of operations required is 6, for example:
$$
\begin{array}{l}
(21,10,11) \rightarrow(23,9,10) \rightarrow(22,8,12) \\
\rightarrow(24,7,11) \rightarrow(23,6,13) \rightarrow(25,5,12) \\
\rightarrow(24,4,14) .
\end{array}
$$

Since the minimum number of stones in one pile decreases from 10 to 4, at least 6 operations are required, so it is impossible to achieve in fewer than 6 operations.
(2) It cannot be achieved.
Since after each operation, the number of stones in each pile either decreases by 1 or increases by 2, we can write it as
$$
(a, b, c) \rightarrow(a-1, b-1, c+2) \text {. }
$$

If the remainders of $a$, $b$, and $c$ when divided by 3 are all different, then the remainders of $a-1$, $b-1$, and $c+2$ when divided by 3 will also be different. Since 21, 10, and 11 have remainders of 0, 1, and 2 when divided by 3, and 14, 14, and 14 all have a remainder of 2 when divided by 3, it is impossible to achieve.