First, right shift by 1 effectively does deleting by 2 with forgetting a remainder. But the remainder could be needed for having the exact result. For instance, change your initial example with adding C000 to 8000, or C002 to 7FFE. Both give the same sum but, sum of shifted values is A000 instead of your 9FFF, and this is definitely more correct. So, you can do such shifting only if sum of LSBs could be lost. In your case with 2 summands and 1 bit shift, this means no more than 1 summand could have 1 in its LSB.
Second, consider this is fixed and you've got A000. A simple ideal math says (a+b)/2 == a/2 + b/2. For your case, the carry bit you initially ignored weighed 0x10000, but after shifting by 1, it weighs 0x8000. That is exactly how A000 differs from your expected 2000. So, if you are sure in other aspects of your method, finish it with logical AND with ~0x8000 == 0x7FFF.