Prove that the sum of two elements is equal third. Please help me solve this problem.
The set {1, 2, 3,..., 3n} is decomposed into three disjoint subsets, with n elements in each. Prove that from each one we can take an element such that the sum of two of the elements is equal to the third. 
 A: I quote from Guy, Unsolved Problems In Number Theory, 3rd edition, problem E11, page 324: 
"Call a partition of the integers $[1,n]$ into three classes admissible if there is no solution to $x+y=z$ with $x,y,z$ in distinct classes. There is no admissible partition with the size of each class $\gt(1/4)n$." 
[Note that $(1/4)n$ is best possible, as one can partition the integers into the odds, the twice-an-odds, and the multiples of 4.]
This is a bit stronger than what the current question asks for, as we have each class being of size $n/3$, which certainly exceeds $(1/4)n$. 
The reference is Esther and George Szekeres, Adding numbers, James Cook Math Notes 4 No. 35 (1984) 4073-4075. This is available here. 
This could be seen as an early result in what has come to be known as "rainbow Ramsey theory", and that search term should lead to more results along these lines. 
EDIT: The problem is also discussed on pages 188-9 of the book Mathematical Miniatures, by Savchev and Andreescu. They present a short solution by V Alexeev, which was published in Kvant, No. 8, 1987. 
