Number theory problem in Israeli Mathematical Olympiad 1995

Problem: The positive integers $d_1, d_2, .. d_n $divide 1995. Prove that there exist $d_i $and $d_j $among them, suach that the numerator of the reduced fraction $d_i$/$d_j $is at least n.

Solution. Note that 3 · 5 · 7 · 19 = 1995. If the chosen divisors include one divisible by 19 and another not divisible by 19, the quotient of the two has numerator divisible by 19, solving the problem since n ≤ 16. If this is not the case, either all divisors are or divisible by 19 or none of them has this property, and in particular n ≤ 8. Without loss of generality, assume the divisors are all not divisible by 19.
Under this assumption, we are done if the divisors include one divisible by 7 and another not divisible by 7, unless n = 8. In the latter case all of the divisors not divisible by 19 occur, including 1 and 3 · 5 · 7, so this case also follows. We now assume that none of the chosen divisors is divisible by 4, so that in particular n ≤ 4.
Again, we are done if the divisors include one divisible by 5 and another not divisible by 5. But this can only fail to occur if n = 1 or n = 2. The former case is trivial, while in the latter case we simply divide the larger divisor by the smaller one, and the resulting numerator has at least one prime divisor and so is at least 3. Hence the problem is solved in all cases