Do there exist distinct integers $p$ and $q$ such that $\dfrac{p}{n} = \dfrac{q}{n+1}$ for some integer $n$, given $p$ is less than $n$ and likewise $q$ is less than $n+1$?
Assume that $\dfrac{p}{n}$ and $\dfrac{q}{n+1}$ can be rational, non-integers.
edit: changed less than inequality constraint to strictly less than.
Your equation can be rewritten as $\frac n{n+1}=\frac pq$, therefore the answer is yes, since $n$ and $n+1$ are necessarily coprime, the fraction $\frac n{n+1}$ is reduced (and the unique integers are the obvious $p=n$, $q=n+1$).
Edit: A direct consequence of the above is that the answer to the new question is no. Since the only (unique!) solution is to have $p$ and $q$ respectively equal to $n$ and $n+1$.