For $N\in \mathbb{N}$, do there exist natural numbers $N$N$ is a natural number. 
Is there any $k$ and some natural numbers $N<n_1<n_2<\cdots<n_k$ such that $$\frac{1}{n_1}+\frac{1}{n_2}+\cdots+\frac{1}{n_k}=1$$?
 A: You may start from:
$$ \frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1 \tag{1}$$
and use the identity:
$$ \frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(n+1)} \tag{2}$$
to increase the number of terms in the LHS of $(1)$. Quoting Wikipedia, any fraction $\frac{x}{y}$ has an Egyptian fraction representation in which the maximum denominator is bounded by $O\left(\frac{y\log^2 y}{\log\log y}\right)$ and a representation with at most $O(\sqrt{\log y})$ terms.

Another possible approach is the following one: 
$$ \frac{1}{N}\prod_{k=N}^{N^2-1}\left(1+\frac{1}{k}\right)=1,\tag{3}$$
and if we expand the LHS of $(3)$ we get a sum of Egyptian fractions with a denominator $\geq N$. 
We just need to remove duplicates: for such a task, we may use $(2)$.
A: Jack D'Aurizio best me to it with a cleaner solution, but I'll post this anyway. We will use the result that,

For any rational $q \in \Bbb Q,$ there exists finitely many $n_1 < n_2 < \dots <n_k$ such that,
  $$ \sum_{i=1}^k \frac1{n_i} = q $$

There are many known algorithms of constructing such an expansion, but in particular there is a simple, greedy one that continuously adds the largest possible unit fraction such that the sum does not exceed $q.$ This is outlined on this page.
Then then let,
$$ q = \sum_{n=1}^N \frac1n + 1 $$
Which will gives $n_1 < n_2 < \dots < n_k$ s.t.
$$ \sum_{n=1}^N \frac1n + 1 = \sum_{n=1}^N \frac1n + \sum_{i=1}^k \frac1{n_i} $$
Which gives the desired result.
