SOLVED

Let $f(k)$ be the maximal value of $n_1$ 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
\[f(k)=(1+o(1))\frac{k}{e-1}?\]

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 $1=\sum \frac{1}{n_i}$.