It turns out to be an extremely tough problem. I didn't solve it within 3hrs. I've heard the outline of the solution, and it's definitely a hard TST problem. And nobody solved it during the contest. This problem was in memory of the first CGMO, which was held in 2002.

Taylor's expansion should help

Hint

Is the answer \[\left\{\frac{1}{2002}, \frac{1}{2005}, \frac{1}{2007}, \frac{1}{2008}, \frac{1}{2011}, \frac{1}{2012}, \frac{1}{2014}, \frac{1}{2017}\right\} \quad \text{and} \quad \left\{\frac{1}{2003}, \frac{1}{2004}, \frac{1}{2006}, \frac{1}{2009}, \frac{1}{2010}, \frac{1}{2013}, \frac{1}{2015}, \frac{1}{2016}\right\}?\]That's what I would guess on first glance, since then the difference $|A - B|$ is on the order of $2002^{-5}$, which seems hard to improve. I don't see a method to rigorously prove this, however.

I created a program and that answer is correct. Don't know how to prove though.

CantonMathGuy wrote: Is the answer \[\left\{\frac{1}{2002}, \frac{1}{2005}, \frac{1}{2007}, \frac{1}{2008}, \frac{1}{2011}, \frac{1}{2012}, \frac{1}{2014}, \frac{1}{2017}\right\} \quad \text{and} \quad \left\{\frac{1}{2003}, \frac{1}{2004}, \frac{1}{2006}, \frac{1}{2009}, \frac{1}{2010}, \frac{1}{2013}, \frac{1}{2015}, \frac{1}{2016}\right\}?\] Yes. The official solution goes like this. Set $X=\{ 1/u:u\in A \},Y=\{1/v:v\in B\}$ Step1)Prove $|X|=|Y|=8$ Step2)Prove $\sum_{k\in X} k=\sum_{k\in Y} k$ Step3)Prove $\sum_{k\in X} k^2=\sum_{k\in Y} k^2$ Step4)Prove $\sum_{k\in X} k^3=\sum_{k\in Y} k^3$ Step5)Prove that the only partition satisfying the above condition is $$\left\{ 2002, 2005, 2007, 2008, 2011, 2012, 2014,2017 \right\} \text{and} \left\{ 2003, 2004, 2006, 2009, 2010, 2013, 2015,2016\right\}$$ And Step4 and Step5 is the main difficulty of this problem.

Can you detail how the solution handles Step 5? I was able to get there but can't find a solution which does not resort to trying many cases.

fattypiggy123 wrote: Can you detail how the solution handles Step 5? I was able to get there but can't find a solution which does not resort to trying many cases. I only get the hint. But I still don't know the solution.

For Step 5, it's equivalent to consider partitions of $0-15$. For $0\leq r \leq 4$, let $x_r$ denote the number of elements in $X$ that are equal to $r \pmod{5}$. Similarly define $y_r$. Using $\sum_{k \in X} k(k-1)(k-2) = \sum_{k \in Y} k(k-1)(k-2)$ and Mod 5, we deduce $x_3-y_3 \equiv x_4-y_4 \pmod{5}$. More gerenally, $x_r-y_r$ is a constant Mod 5. Since $x_0+y_0=4$, $x_1+y_1= \dots = x_4+y_4=3$, essentially the only possiblity is $x_0=4, x_1=\dots=x_4=1$, $y_0=0, y_1=\dots=y_4=2$. Thus $0, 5, 10, 15 \in X$. The rest should be easy.

Can anyone post the complete solution,moreover are the steps required to prove the above lemmas algorithmic??Step1 can be proved by contradicting minimality after exchanging numbers in the set A and B.

why mod5?

I think it can be generalized to $2^k$, so for a set $\{n, n+1, ..., n+2^k -1\}$, the desired sets satisfying the condition holds the same pattern

Solution would follow this way. Call a partition of $\{\frac{1}{n},\frac{1}{n+1}, ..., \frac{1}{n+2^k -1}\}$ wonderful if the difference between the sum of two sets are minimum of the whole cases. We'll prove that for each $k$, the pattern continues as choosing the original and inverse of $k-1$. Inverse means for $n$, $n'$,($n'=n+2^{k-1}$) if a set of $\{n, n+1, \cdot\cdot\cdot,n+2^k -1\}$ chose a number in a pattern, for $n'$, choosing rule becomes total opposite. for $k=1$, it is apparent. for $k=2$, the minimum partition is apparent to be $\{\frac{1}{n}, \frac{1}{n+3}\}$ and $\{\frac{1}{n+1}, \frac{1}{n+2}\}$, which is true to choose the original and inverse of the pattern. So at first it is easy to show that partitions have the same size. Then we need to show that partition holds numbers in the rule stated for $n, n+2^i, n+2^{i+1}, n+3 \cdot2^i, \cdot\cdot\cdot$)