Are there $4$ sets such that the sum of the two numbers are equal? For each set of $117$ different $3-$ digit natural numbers,can we choose $4$ disjoint sets with $2$ elements $A,B,C,D$ with the identity:the sums of the two numbers of each set are equal?
How can I check this??
I got stuck right now...
Could you give me a hint??
 A: From the requirement that each of the set $A,B,C,D$ should have two elements, I presume that sum of two elements of a set means 'sum of two distinct elements of a set'. Otherwise, the solution below can be modified accordingly.
The least possible sum is 100+101=201 and maximum possible sum is 998+999=1997. But, not every sum is expressible as sum of four distinct unordered pairs. For example, 205=100+105=101+104=102+103 -- only three pairs. The least sum that is expressible as sum of four distinct unordered pairs is 207(=103+104) and the maximum is 1991(=995+996). Hence the number of sums that are expressible as sum of at least four distinct unordered pairs is 1991-206=1785.
Let $S$ be the multiset of all possible sums of 117 distinct three digit numbers. The following sums can occur with their maximum multiplicities.

sum -- multiplicity
  
  
*
  
*201 -- 1
  
*202 -- 1
  
*203 -- 2
  
*204 -- 2
  
*205 -- 3
  
*206 -- 3
  
*207 -- 4 or 4+
  
*...
  
*1991 -- 4 or 4+
  
*1992 -- 3
  
*1993 -- 3
  
*1994 -- 2
  
*1995 -- 2
  
*1996 -- 1
  
*1997 -- 1
  

Hence, there are $\binom{117}{2}-24=6774$ sums expressible as sum of at least four distinct unordered pairs. We have $\frac{6774}{1785}=3.79$ (up to two digits). By pigeon hole principle, there must be four distinct unordered pairs with the required property. Among the four distinct ordered pairs, we cannot have unordered pairs of form $(a,b)$ and $(a,c)$ where $c\ne b$ as $a+b\ne a+c$. Hence for any two unordered pairs (four distinct ordered pairs), $(m,n)$ and $p,q$, we have $m\ne p,m\ne q,n\ne p,n\ne q$. Hence, we have four sets of size 2, each containing the elements of four unordered pairs respectively.
