A problem based on pigeonhole Numbers 1 to 1994 are divided into 6 sets.Show that at least in one  group there will be two numbers whose sum is also in that group ?
We can prove that at least one group will contain more than 332 elements.
A set is called sum free if it DOES NOT  CONTAIN sum of any two of its elements or does not contain twice of an element
If a set contains natural numbers from 1 to $2n+1$ its sum free subsequence will contain $n+1 $elements.Or in other word if we take the last half numbers $(\frac{n}{2} $ or $\frac{n+1}{2}$ according to number of elements of the set) it will be free.
But I cannot proceed further because my selection need not be consecutive numbers. 
I have to prove that if a set contains more than 332 numbers it cannot be sum free..But How..?
 A: This is from IMO 1978. The original problem is
An international society has members from six different countries. The list of members contains 1978 names, numbered 1, 2, . . . , 1978. Prove that there is at least one member whose number is the sum of the numbers of two members from his own country or twice as large as the number of one member from his own country.
Here is the solution from Arthur Engel's Problem Solving Strategies, which is where I first saw it

A: Schur's theorem (one of many things called that, at least) states that partitioning $\{1, ..., n\}$ into $k$ subsets $S_1, \dots, S_k$ always has some $S_i$ not sum-free (i.e., there exist $x, y\in S_i$ with $x + y\in S_i$) if $n > k!\, e$. Since $6! e = 1927.16 < 1994$, your result follows. 
Since this is a homework problem, I don't know how much combinatorics you've covered already and what tools are available to you; as such, I'd recommend you look up proofs of Schur's theorem if it hasn't been presented in class already and adapt the arguments to this case. (I don't know the proof offhand, but it seems to be amenable to standard Ramsey-style arguments. The value of $1994$ in the original problem is also suspiciously close to the bound from Schur's theorem.)
