12. There are a total of 65 numbers written on the blackboard, including 11 ones, 12 twelves, 13 thirteens, 14 fourteens, and 15 fifteens. Each time, the operator erases 4 different numbers and writes down one of the fifth kind (for example, erasing 11, 12, 13, 14, and writing a 15, or erasing 12, 13, 14, 15, and writing an 11...). If after several operations, there are exactly two numbers left on the blackboard, these two numbers are $\qquad$.
【Answer】12, 14
【Analysis】Key Point: Parity
Consider the parity: There are 3 odd numbers on the blackboard, and at the beginning, the differences in the number of each pair are even. Each operation may result in the following changes: (1) Each of the 3 numbers decreases by 1, so the differences between each pair remain unchanged; (2) One number increases by 1, and the other two numbers each decrease by 1, so among the three differences between each pair, one remains unchanged, and two increase or decrease by 2. In summary, the parity of the differences between the three numbers never changes, and they are always even. It is easy to see that after the operations, at least one of the three odd numbers will be reduced to 0. If there is an odd number among the two remaining numbers, the difference in the number of these two odd numbers will be 1, which is not even. From the above discussion, we can conclude that there are no odd numbers among the remaining numbers.
There are 2 even numbers on the blackboard, and at the beginning, the difference in their numbers is 2. Each operation may result in the following changes: (1) Both numbers decrease by 1, so the difference remains unchanged; (2) One number increases by 1, and the other decreases by 1, so the difference increases or decreases by 2.
Each operation reduces 3 numbers on the blackboard, so the total number of operations is $(65-2) \div 3=21$ times. In these 21 operations, all 39 odd numbers disappear. For each operation involving odd numbers, 1 odd number is reduced; for each operation involving even numbers, 3 odd numbers are reduced. Therefore, in these 21 operations, 12 operations involve odd numbers, and 9 operations involve even numbers. Thus, considering the difference in the number of the two even numbers, the difference remains unchanged in the 12 operations involving odd numbers, and changes by 2 (either increases or decreases) in the 9 operations involving even numbers, a total of 9 times, which is an odd number of times. Therefore, after the operations, the difference in the number of the two even numbers will change. At the beginning, the difference is 2, and obviously, only two even numbers remain at the end. If these two even numbers are the same, the difference is 2; if they are different, the difference is 0. Considering that the difference was 2 at the beginning and changes at the end, the two remaining even numbers on the blackboard are different, and they are 12, 14.