I had this question bothering me for a while, but I can't come up with a meaningful answer.
The problem is the following:
Let integers $a_i,b_j\in${$1,\ldots,n$} and $K_1,K_2\in$ {$1,\ldots,K$}, then how small (as a function of $K$ and $n$), but strictly positive, can the following absolute difference be.
$\biggl|(\sum_{i=1}^{K_1} \frac{1}{a_i})-(\sum_{j=1}^{K_2} \frac{1}{b_j})\biggr|$
As an example for $K_1=1,$$K_2=1$ choosing $a_1=n,$$b_1=n-1$gives the smallest positive absolute difference, that is $\frac{1}{n(n-1)}$. What could the general case be?
Best Answer
A sum of $K$ unit fractions, each of denominator $n_i \leq n$, can be rewritten as a fraction with a denominator bounded by the product of the $n_i$, i.e. by $n^K$. (A small improvement is possible, with product of distinct integers $\leq n$)
A difference of two such fractions
(which are themselves sums of $K_1$ and $K_2$ fractions, respectively), is a fraction with a demominator bounded by $n^{K_1+K_2}$. (If $K_1\neq K_2$ the order of magnitude of the difference is actually $\frac{1}{n^{\min(K_1,K_2)}}$.)
Let us assume that $K_1=K_2$. So, the smallest nonzero difference $d(K,n)$ of sums of $K$ unit fractions, with $n_i \leq n$ is $d(K, n)\leq \frac{1}{n^{2K}}$.
I believe that this order of magnitude, with slightly weaker constants, can be achieved with a constructive parametrization.
Example: If $K_1=K_2=2$, then choose $\frac{1}{x^2 + 4 x + 1} + \frac{1}{x^2 + 4 x + 3} - \frac{1}{x^2 + 3 x + 1} - \frac{1}{x^2 + 5 x + 5}= \frac{2}{(1 + 3 x + x^2) (1 + 4 x + x^2) (3 + 4 x + x^2) (5 + 5 x + x^2)}$. Now, taking $n=x^2+5x+5$, one has a set of 4 unit fractions, for which the difference of the sums above is asymptotically $ \frac{2}{n^4}$. This example is (for $K_1=K_2=2$) possibly the best one can find, (but I did not prove this).
I conjecture that for larger values of $K$ one can construct similar polynomial examples.
Quite possibly this has applications in questions in diophantine approximation, exponential sums, large sieve etc.