For certain pairs $(m,n)$ of positive integers with $m\geq n$ there are exactly $50$ distinct positive integers $k$ such that $|\log m - \log k| < \log n$. Find the sum of all possible values of the product $mn$.
We have $\log m - \log k = \log \left( \frac mk \right)$, hence we can rewrite the inequality as follows:
\[- \log n 1$ the left inequality can be rewritten as $m\leq\dfrac{51n}{n^2-1}$, and the right one as $m > \dfrac{50n}{n^2-1}$.
Remember that we must have $m\geq n$. However, for $n\geq 8$ we have $\dfrac{51n}{n^2-1} < n$, and hence $m<n$, which is a contradiction.
This only leaves us with the cases $n\in\{2,3,4,5,6,7\}$.

For $n=2$ we have $\dfrac{100}3 < m \leq \dfrac{102}3$ with a single integer solution $m=\dfrac{102}3=34$.
For $n=3$ we have $\dfrac{150}8 < m \leq \dfrac{153}8$ with a single integer solution $m=\dfrac{152}8=19$.
For $n=4,5,6,7$ our inequality has no integer solutions for $m$.
Therefore the answer is $34\cdot 2 + 19\cdot 3 = 68 + 57 = \boxed{125}$.