SOLVED

Let $f(k)$ be the minimal value of $n_k$ such that there exist $n_1<n_2<\cdots <n_k$ with
\[1=\frac{1}{n_1}+\cdots+\frac{1}{n_k}.\]
Is it true that
\[f(k)=(1+o(1))\frac{e}{e-1}k?\]

It is trivial that $f(k)\geq (1+o(1))\frac{e}{e-1}k$, since for any $u\geq 1$
\[\sum_{e\leq n\leq eu}\frac{1}{n}= 1+o(1),\]
and so if $eu\approx f(k)$ then $k\leq \frac{e-1}{e}f(k)$. Proved by Martin [Ma00].