Can $\sum_{k\in M}\frac{1}{k}$ be a large integer? I'm interested in the following question:
Given an integer $n_0$. Is there always an integer $n>n_0$ and a finite subset $M\subset \mathbb N$ with $\sum_{k\in M}\frac{1}{k} = n$.
This is not a homework problem, I don't know if there is an easy solution.
I appreciate any hint.
 A: Yes, and more is true: given any positive rational number $n$, there exists a finite set $M\subset\Bbb N$ such that $\sum_{k\in M} \frac1k = n$.
Perhaps the most straightforward proof is this: since the harmonic series diverges, there exists a unique $m\in\Bbb N$ such that
$$
\sum_{k=1}^m \frac1k \le n < \sum_{k=1}^{m+1} \frac1k.
$$
(If $n$ is small then $m$ might equal $0$.)
Write $r=n-\sum_{k=1}^m \frac1k$, which is a rational number less than $\frac1{m+1}$. Then use the greedy algorithm to write $r$ as an Egyptian fraction $\sum_{k\in M_r} \frac1k$. By size considerations, every element of $M_r$ exceeds $m$, and so $M=\{1,\dots,m\} \cup M_r$ has the property that $\sum_{k\in M} \frac1k = n$.
Similar constructions can yield representations of $n$ with particular constraints; for example, one can choose any $j\in\Bbb N$ and force all the elements of $M$ to exceed $j$.
A: Clearly for any $n_0$ there exists a (not necessarily integer) $N$ and a set $S$ such that $n \le N < n + 1 $ by the divergence of the harmonic series with:
$$ N = \sum_{k \in S} \frac1k $$
Let $ q = n + 1 - N $. It is well known that there exists a finite decomposition of $q$ into unit fractions with unique denominators (call the multiset of such $Q$). Using the identity:
$$ \frac{1}{a} = \frac{1}{a+1} + \frac{1}{a(a+1)}$$
we can make the denominators in $Q$ larger than the largest element of $S$. (We have to be careful how we do this, but clearly it is possible.) Since $Q$ is finite, this process terminates. Now $Q \cup S $ is the desired set with $\sum_{k \in Q \cup S} \frac1k = n + 1$, and $Q \cap S = \emptyset$.
A: The Erdős–Graham conjecture in combinatorial number theory states that, if the unit fractions are partitioned into finitely many subsets, then one of the subsets has a finite subset of itself whose reciprocals sum to one.
That would imply that we can write every positive integer as sum of unit fractions: We show that we can construct a sequence $(A_n)_{n\in\mathbb N}$ of disjoint finite subsets of $\mathbb N\setminus \{1\}$ such that $\sum_{i\in A_n} \frac{1}{i}=1$ for all $n\in\mathbb N$. We construct the sequence by induction: Let $n\ge 1$ be given and $A_1 ,\dots A_n$ be already defined. We split each of those $A_i$ in two non-empty set and define
$$B:=\mathbb N \setminus \left( \{1\} \cup \bigcup_{i=1}^n A_i \right).$$
Then those $2n+1$ sets are a partition of $\mathbb N\setminus\{1\}$. By the stated conjecture $B$ contains a finite subset $B'$ such that $\sum_{i\in B'} \frac{1}{i}=1$. We define $A_{n+1}:=B'$.
In order to write a given positive integer $n$ as sum of unit fractions we simply define
$$S:= \bigcup_{i=1}^n A_i $$
and have 
$$\sum_{i\in S} \frac{1}{i}=n.$$
A: According to this paper there are even infinitely many numbers $n$ that can be written as sum of unit fractions. It provides a divergent lower bound on the number $|N(n)|$ of integers that can be written as the sum of unit fractions whose denominator is at most $n$.
Another paper on this topic can be found here. It states that the splitting algorithm terminates for every positive rational. This means that every positive integer admits a representation as the sum of unit fractions.
These papers show that the result you're asking for is true, but I don't know if there is any elementary proof of this theorem.
