Answer: 125

Small Hint:

The inequality says exactly that $\frac{m}{n} \lt k \lt mn$

Big Hint:

Write $m = nq + r$ with $0 \le r \lt n;$ the count of such $k$ is $mn - q - 1 = 50,$ so $q(n^2 - 1) + nr = 51$ leaves only a few cases

Solution:

The inequality $|\log m - \log k| \lt \log n$ is equivalent to $\frac{m}{n} \lt k \lt mn.$ Write $m = nq + r$ with $0 \le r \lt n;$ since $m \ge n \ge 2$ (for $n = 1$ no $k$ works), $q \ge 1.$ The integers $k$ in the interval are $q + 1,$ $q + 2,$ $\ldots,$ $mn - 1,$ so there are $mn - q - 1 = 50$ of them, that is, $mn - q = 51,$ or $$q(n^2 - 1) + nr = 51.$$

For $n \ge 8$ the left side is at least $63,$ so $2 \le n \le 7.$ Checking each case, only $n = 2,$ $r = 0,$ $q = 17$ (so $m = 34$) and $n = 3,$ $r = 1,$ $q = 6$ (so $m = 19$) work. These give $mn = 68$ and $mn = 57;$ indeed $17 \lt k \lt 68$ and $\frac{19}{3} \lt k \lt 57$ each contain exactly $50$ integers.

The sum of all possible values of $mn$ is $68 + 57 = 125.$