Reciprocal sum with 2017 Find 2017 distinct postive integers ($a_1, a_2, \dots, a_{2017}$) such that
$$\sum_{i=1}^{2017}\dfrac{1}{a_i}=\dfrac{2017}{1000}$$
That is what I know: $1=1/2+1/3+1/6$. From that we can find infinity many ways to reach 1 as sum of reciprocals of distinct positive integers.

You can also start with 
$$ \frac{2017}{1000}=\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{5}+\frac{1}{15}+\frac{1}{3000} $$
or
$$ \frac{2017}{1000}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+\frac{1}{60}+\frac{1}{3000} $$
then expand $\frac{1}{3000}$ into a sum of $2012$ egyptian fractions by exploiting $\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(n+1)}$ multiple times, or
$$\frac{1}{3000}=\frac{1}{3000}\underbrace{\left(\sum_{k=1}^{2010}\frac{1}{2^k}+\frac{1}{3\cdot 2^{2009}}+\frac{1}{6\cdot 2^{2009}}\right)}_{2012\text{ terms}}$$
The number of possible solutions for this problem is probably gargantuan.
 You know I've always liked that word gargantuan? I so rarely have an opportunity to use it in a sentence.
