Solution:

The answer is (C). The numbers must be either all even or all odd, since the sums taken two at a time are all even.

Similarly, the numbers must all be divisible by 3, because if $A$ contained a number not divisible by 3 (with remainder $r$, say), then there could be only one other such number (with remainder $3 - r$) and hence $|A| = 2$, too few!

Now the even multiples of 3, that is, the multiples of 6 between 1 and 100, are 16, while the odd multiples of 3 are 17, so $A = \{3(2k+1) \mid k = 0, 1, \ldots, 16\}$.