So, it is required to prove that for each natural $k$ there are two natural numbers with a difference $k$ having the same number of divisors. For example, for the case $k=27$, the pair $(18,45)$ is suitable.
One of the special cases of this problem can be solved quite easily, namely, if in the factorization of the number $k$ into factors there is an equal number of twos and triples (in particular, if there are none at all), then the numbers $2k$ and $3k$ will have the same number of divisors.
Please help to solve this problem in the general case.