Harmonic differences

Do there exists two positive integers $h$ and $l$ such that $\sum_{i = l}^h \frac{1}{i}$ is an integer other than the trivial case?

Every natural number can be written as $2^\alpha x$. More specifically for each number $n$, $\alpha$ is the n-th element of this sequence, and $x$ is the n-th term of this sequence.
Let us first have a look at $\alpha$. If you take a consecutive subset, from index $l$ to $h$, the maximum always appears ones.

Proof:

Assume the maximum $y$ appears at least twice. As the index of the number has to be of the form $2^\alpha x$, the two appearances can be assumed to be $2^y a$ and $2^y b$ with $a$ and $b$ odd and $a < b$. As $2^y a < 2^y (a+1)< 2^y b$, and both ends are in the consecutive subset, $ 2^y (a+1)$ has to be in the subset. But as $a$ is odd, $a+1$ is even, and such $a+1$ can be written as $2^k z$ with k nonzero and $z$ odd. Hence, $2^y (a+1) = 2^{y+k} z$, so there exists a value $y+k$ larger than $y$ for the exponent, so $y$ was not the maximum, which leads to a contradiction.

Now, take the sum $\sum_{i = l}^h \frac{1}{i}$. By the previous lemma, there is one single element that contains the largest power of two, if written like $2^\alpha x$. Remove that element from the sum. The sum of all other elements can be written as $\frac{p}{\text{lcm} (i)}$ with $a$ an integer and $\text{lcm} (i)$ the least common multiple of all denominators. As all of the denominators have a lower exponent for two than $\alpha$, $\text{lcm} (i)$ is not a multiple of $\alpha$. Let us write this is $2^\beta y$ with $\beta < \alpha$

Our total sum is now $\frac{1}{2^\alpha x} + \frac{p}{2^\beta y} = \frac{2^\beta y + 2^\alpha x p}{2^\alpha x 2^\beta y}$
As the denominator is a multiple of $2^\alpha$ and the numerator is not, this can never be an integer. QED