This question is from RMO 2017, the second round of selection for the IMO in India.
Question: Prove that there cannot exist a strictly increasing infinite harmonic progression consisting of terms which are positive integers.
My proof:
Let the first term of a HP is a positive integer $a$. Suppose
$$\frac{1}{a}, \frac{1}{a}+d,\frac{1}{a}+2d,\dots (d \in \mathbb R)$$ or
$$\frac{1}{a}, \frac{1+ad}{a},\frac{1+2ad}{a},\dots$$
is an infinite AP such that the reciprocals of its terms form a strictly increasing HP consisting of positive integers. So the HP becomes
$$a,\frac{a}{1+ad},\frac{a}{1+2ad},\frac{a}{1+3ad},\dots$$
Since the term $\frac{a}{1+ad}$ is a positive integer, $1+ad|a$. This implies that $1+ad$ is a factor of $a$.
Similarly $1+ad, 1+2ad, 1+3ad, \dots$ are all factors of $a$. But it is impossible for a number to have infinite factors unless all of $1+ad, 1+2ad, 1+3ad, \dots$ are equal. That is
$$1+ad=1+2ad=1+3ad=\dots$$ or
$$ad=2ad=3ad=\dots$$
$$\implies ad=0$$
Since a is a positive number,
$$d=0$$
But if $d=0$ then all the terms of the HP become 1, which contardicts the hypothesis that it is strictly increasing. Hence, no such HP exists.
For some reason, I think there's a flaw in my proof. Is my proof correct? And even if it is, can it be used as a rigorous proof in a contest?
This proof is indeed flawed. The statement that a number can't have infinitely many factors is wrong when it comes to $\mathbb R$; for instance, $1=n(\frac{1}{n})$ is a trivial factorisation of $1$ in infinitely many ways.
What you want to do is show that eventually $|1+nad|>|a|$ (assuming $d\neq 0$), so $$\left|\frac a{1+nad}\right|<1.$$
I'll leave this as an exercise.