This is a beautiful puzzle from Mathematical Circles (Russian Experience) by Dmitri Fomin, Sergey Genkin and Ilia Itenberg.
There numbers 1, 2, 3, ….. , 1989 are written on a blackboard. It is permitted to erase any two of them and replace them with their difference. Can this operation be used to obtain a situation where all the numbers on the blackboard are zeros?
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Happy zeroing in on the solution!
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Proof by construction:
Operation applied on 1 and 2 give 1 and 1. And then applied on 1 and 1 gives 0 and 0.
Now for any number x,
Operation applied on 0 and x give x and x. And then applied on the two numbers gives 0 and 0.
Apply the operation on all the numbers and get zeroes. Done.
Solution seems too simple though. May be I understood the question wrong?
If you choose two numbers A and B, you erase them and put |A-B| in their place. So every time you do it, you effectively reduce one count from the board.
Aah. My bad. Misunderstood the question. The solution is so elegant and beautiful. Thanks
Let the numbers be a, b, c ….
For the final sum to be zero, all the numbers will belong to one of the two groups – group 1 or group 2 such that ( sum (group 1)) – (sum (group 2)) = 0
For example, lets the number be 6,7,8,9 then 6 and 9 will belong to group 1 and 7 and 8 to group 2.
Now, sum(group 1 & 2) = 2* sum (group 1) which implies sum(group 1 & 2) = even number
Sum of (1,2…., 1989) is an odd number, therefore it is not possible to get 0 finally.