TITLE: For $N\in \mathbb{N}$, do there exist natural numbers $N<n_1<n_2<\cdots<n_k$ such that $\frac{1}{n_1}+\cdots+\frac{1}{n_k}=1$?
QUESTION [16 upvotes]: $N$ is a natural number. 
Is there any $k$ and some natural numbers $N<n_1<n_2<\cdots<n_k$ such that $$\frac{1}{n_1}+\frac{1}{n_2}+\cdots+\frac{1}{n_k}=1$$?

REPLY [11 votes]: 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.