Let $\mathbb{N}$ denote the set of positive integers. If $A\subseteq \mathbb{N}$ is finite, we say that $A$ is *reciprocally summable to $1$* ("rs1") if $\sum\_{a\in A} \frac{1}{a} = 1$.


If $A\subseteq \mathbb{N}$ is finite and $\sum\_{a\in A} \frac{1}{a} < 1$, is there a finite rs1 set $A'$ with $A\subseteq A'$?



On subsets of $\mathbb{N}$ reciprocally summable to $1$
Yes, several algorithms for Egyptian fractions suggested by Gerhard Paseman works. 


We want to represent the number $r=1-\sum\_{a\in A}$ as a sum of distinct Egyptian fractions with denominators not in $A$. This may be done by many ways, for example we may use


Lemma. For any positive integers $a,n$ the number $1/a$ is representable as a sum of distinct Egyptian fractions with denominators greater than $n$.


Proof. Start with $1/a=1/a$. If it does not work (i.e., $a\leqslant n$), replace $1/a$ to $1/(a+1)+1/(a^2+a)$. After that do the same with both $1/(a+1),1/(a^2+a)$. We get four fractions which sum up to $1/a$, then eight fractions and so on. Stop when we get, say, $2^k$ fractions which are all less than $1/n$. They are all distinct (unless $a=1$, in this case use $1=1/2+1/3+1/6$), that proves Lemma. Why distinct? Assume that two of them coincide, choose the first step when this happens, say $b+1=c^2+c$ where $b,c$ were denominators on the previous step. Then the total number of steps is less than $c$, and $b=c^2+c-1$ was obtained from $c^2+c-2$, this guy in turn from $c^2+c-3$ and so on. Thus initial number $a$ was not less than $c^2$ and could not generate $c$. Lemma is proved.


Now start with the representation $r=1/N+\ldots+1/N$ for large $N>\max(A)$. Then perform the replacement algorithm: while two fractions $1/a,1/a$ in the representation are equal, replace one of them using lemma by a sum with denominators greater than anything already used. After finitely many steps we remove all repetitions.


