Let $N(m)$ be the max of $n$ such that $\sum_{i=1}^n\frac{1}{a_i}=1$
where $a_i \ (i=1,2,\cdots,n)$ are integers which satisfy $2\le a_1\lt a_2\lt\cdots\lt a_n\le m$.
Question 1 : What is $N(99)$?
Question 2 : What is $N(m)$?
Examples : I'm going to represent $\sum_{i=1}^n\frac{1}{a_i}$ as $(a_1,a_2,\cdots,a_n).$
The followings are two examples for $(m,n)=(99,42)$.
$(15,17,20,21,22,26,27,30,32,33,34,35,36,38,39,40,42,44,45,48,50,52,54,55,56,60, 63,66,70,75,76,77,78,80,84,85,88,90,91,95,96,99)$
$(17,18,20,21,22,24,26,27,32,33,34,35,36,38,39,40,42,44,45,48,50,52,54,55,56,60,63,66,70,72,75,76,77,78,80,84,85,88,91,95,96,99)$
Remark :
Question 1 has been asked on math.SE.
$99$ has no special meaning except that $99$ is not too small and not too large.
Question 2 might be somewhat ambiguous. The best answer would be to represent $N(m)$ by $m$ if it is possible. Also, finding both the max of $m$ and the min of $m$ would be needed.
Motivation : The beginning was the following:
"$\sum_{k=2}^n \frac 1k$ is not an integer for any $n$."
(the proof and the other related facts can be seen at https://math.stackexchange.com/questions/494174/proving-that-the-finite-sum-of-the-each-reciprocal-of-any-sequence-of-integers-w).
Best Answer
Note that $N(m)$ cannot be significantly larger than $m(1-1/e)$, since the sum of the reciprocals of any $m(1-1/e)$ integers not exceeding $m$ is at least $\sum_{m/e < k < m} 1/k \sim 1$.
I've proved that in fact, $N(m)$ is asymptotic to $m(1-1/e)$; in other words, you can have essentially as many summands as size considerations allow you to have. See Theorem 1 and equation (5) of my paper "Denser Eygptian fractions". (There I consider the equivalent dual problem - instead of fixing $m$ and trying to maximize the number of summands, I fix the number of summands and try to minimize $m$.)