Let $2^m$ be the highest power of $2$ that does not exceed $n$. Then none of the other positive integers less than or equal to $n$ are divisible by $2^m$. Let $k = \text{lcm}(1, 2, \dots, n)$. Now write each of the terms $1, \frac{1}{2}, \dots, \frac{1}{n}$ as fractions with denominator $k$. All will have even numerators except $\frac{1}{2^m}$ which will have an odd numerator. Thus their sum is a fraction $\frac{h}{k}$ with $h$ odd, so it cannot be an integer.