Integers as a sum of $\frac{1}{n}$ Say $\sum_{i \in I} \frac{1}{n_i} = 2$, where $(n_i)_{i \in I}$ is a finite sequence of positive integers (not necessarily distinct). Is there a subsequence $(n_i)_{i \in J}$ of $(n_i)_{i \in I}$ such that $\sum_{i \in J} \frac{1}{n_i} = 1$?

Here is a counterexample even when all fractions are required to be different!
$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13}+\frac{1}{16}+\frac{1}{17}+\frac{1}{21}+\frac{1}{1105}+\frac{1}{55692}+\frac{1}{1361360}$
This came from:
$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13}+\frac{1}{16}+\frac{1}{17}$ [only prime powers]
$+\frac{1}{3 \cdot 7}+\frac{1}{5 \cdot 13 \cdot 17}+\frac{1}{2^2 \cdot 3^2 \cdot 7 \cdot 13 \cdot 17}+\frac{1}{2^4 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17}$ [plus extra terms to get to $2$]
To prove it, first multiply throughout by $2^4 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17$.
Now consider any subset whose sum is $1$. We can assume that it does not include the last term. Modulo $13$ we see immediately that it cannot include any term with denominator divisible by $13$. Now modulo $7$ we see that it cannot include $\frac{1}{21}$. We can repeat this reasoning all the way, but now actually it is obvious that it is impossible because the remaining terms allowed are all prime powers and hence the term in the subset with largest denominator $p^k$ will create a contradiction modulo $p$.
