This is a Multiple Choice Question :(

Consider only the numerators, since the denominators are all equal to 1000. For the sum of the elements of a sub-multiset to be an integer, the sum of the numerators must be $\equiv 0 \pmod{1000}$ .

Now, consider $S_1, S_2, S_3 \ldots S_{1000}$ where $S_i$ denotes the sum of the numerators of the first i elements of A. If one of them is divisible by 1000, we are done, or else, their remainders modulo 1000 must be from the set $[1,2,3, \ldots 997,998,999]$ .

Since there are 999 possible remainders and 1000 $S_i$ 's, by PHP, the remainders of two $S_i$ 's must be the same. Let them be $S_j$ and $S_k$ .

Therefore $S_j -S_k \equiv 0 \pmod{1000}$ . Or, alternatively, we can say that the set $[j+1,j+2 \dots k-1,k]$ satisfies the condition

It's simple really. The numerators must be all rational numbers , and it is always possible to get an integer out of rational numbers [even fractions] by multiplying [multi sub set] and adding. For example 3.1416 repeated 10,000 times gives us 31416 , an integer.