Does $k(N)-(e-1)N\to\infty$? Terms needed for a unit-fraction sum to $1$ with denominators $\geq N$ (Erdős #295)

For $N\geq 1$ let $k(N)$ be the smallest $k$ for which there exist integers $N\leq n_1<\cdots<n_k$ with $$1=\frac1{n_1}+\cdots+\frac1{n_k}.$$ That is, $k(N)$ is the minimum number of terms in a representation of $1$ as a sum of distinct unit fractions all of whose denominators are at least $N$. Is it true that $$\lim_{N\to\infty}\bigl(k(N)-(e-1)N\bigr)=\infty?$$

Acceptance. FULLY RESOLVES (proof-shaped): a complete proof that $k(N)-(e-1)N\to\infty$ as $N\to\infty$, or a proof that it remains bounded (i.e. $\limsup_{N}\,[k(N)-(e-1)N]<\infty$), with rigorous justification. A machine-checkable proof is preferred, otherwise a full written proof; a computation alone cannot settle the limit. ADVANCES (each independently checkable): (a) improve the Erdős–Straus lower bound $k(N)-(e-1)N>-c$ to a divergent lower bound $k(N)-(e-1)N\geq g(N)$ with $g(N)\to\infty$, with proof; (b) improve the upper bound $k(N)-(e-1)N\ll N/\log N$, with proof; (c) compute $k(N)$ exactly over a new range of $N$ via a certified shortest-representation search, extending OEIS A192881 and exhibiting the behaviour of $k(N)-(e-1)N$. Deliver the proof, or the search code plus the certified $k(N)$ table.