Proof: consider the sum S = 1/1 + 1/2 + ... + 1/n. Suppose 2^k is the biggest power of 2 that's <= n. Then one of the terms in S is 1/2^k, and there is nothing else with a multiple of 2^k in the denominator. So when you clear denominators, there'll be exactly one term (that one) with an odd numerator, so the whole thing is (odd number) / (multiple of 2^k) and therefore not an integer.