Assume that $f$ is a function such that $\lim_{\varepsilon \rightarrow 0^+} f(\varepsilon) = 1$. Is there an $f$ such that
$$\lim_{\varepsilon \rightarrow 0^+} \lim_{m \rightarrow \infty} \sum_{n=1}^m \frac{1}{n} f(n \varepsilon)$$
exists? In another question, I found an $f$ such that
$$\lim_{\varepsilon \rightarrow 0^+} \lim_{m \rightarrow \infty} \sum_{n=1}^m n^s f(n \varepsilon) = \zeta(-s)$$
for $s \neq -1$, namely
$$f(x) = \mathrm{e}^{-x}\left(1-\frac{x}{s+1}\right)$$
This regulator was obtained by seeking an $f$ such that
$$\int_0^\infty x^s f(x) \,\mathrm{d}x$$
for $s > -1$. Hence we can similarly ask whether there is an $f$ such that
$$\int_0^\infty \frac{1}{x} f(x) \,\mathrm{d}x = 0$$
or whether no such $f$ exists. Here I describe a way of regularizing the harmonic series that is based on the above regulator, but does not take the same form.
I don't have the reputation to comment, so I'll mark my answer as community wiki since it's not a full answer:
I'm not aware of a regulator of exactly the type you are looking for, but the harmonic series is logarithmically divergent, and can be regulated by a kind of "minimal subtraction" scheme to obtain the Euler-Mascheroni constant as a finite answer.
It may be possible to modify this to form a function of the type you want.