Using Statement 1 alone, we might have some scattered list of 20 distinct integers, say 10, 20, 30, 40, ...., 200. If we add one to any individual value in this list, we still have 20 distinct integers. So we can have a list with no consecutive integers.

But our list could also be:

2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3

Then if we add one to any value in the list, we still have only two distinct values in our list, but this list does contain two consecutive integers. So Statement 1 is not sufficient.

Statement 2 is clearly insufficient alone. Combining the two Statements, we know we have a repeated value in the list. Say that value is '2' (or you could use 'x', it won't matter). So we know we have these values in our list:

..., 2, 2, ...

If we now add 1 to one of our repeated values, we get this list:

..., 2, 3, ....

but this list has, according to Statement 1, the same number of distinct values as the previous list. So the '3' we just created cannot be a brand new value that was not previously in the list, and our original list must contain both 2 and 3, so must contain two consecutive integers. So the two Statements together are sufficient and the answer is C.