The upper bound $f(k) \leq (1+o(1))\frac{k}{e-1}$ is trivial since for any $u\geq 1$ we have\[\sum_{u\leq n\leq eu}\frac{1}{n}=1+o(1),\]and hence if $f(k)=u$ then we must have $k\geq (e-1-o(1))u$.

Essentially solved by Croot [Cr01], who showed that for any $N>1$ there exists some $k\geq 1$ and\[N<n_1<\cdots <n_k \leq (e+o(1))N\]with $1=\sum \frac{1}{n_i}$.

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