I seem to be enamoured with this book at the moment, and hence another one from A Moscow Math Circle – Week by Week Problem Sets by Sergey Dorichenko.
Is it possible to write more than 50 two digit numbers on a blackboard without having two numbers on the board whose sum is 100?
As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.
Happy century!
Alok Goyal's Puzzles Page 
The problem can be changed to:
Is it possible to have 50 numbers from [-40, 49] such that sum of no two numbers is 0.
[41,49] would not matter here
So, Is it possible to have 41 numbers from [-40, 40] such that sum of no two numbers is 0.
Not possible.
Alok,
I asked my son Anahat to attempt this question. He mentioned that there are three sets of numbers which need to be ruled out in the very beginning. 1 through 9, 50, and 91 though 99. That leaves us with two sets of numbers. 10-49 and 51-99. Now, in order to avoid 100, we choose from all odd numbers from the first set (10 – 49) and all even numbers from the set (51-99) or reverse. This leads to a total of 1 (number 50) + 9 (from 91 through 99) + 20 (odd selection )+ 20 (even selection). This adds up to 50 numbers. Any additional number chosen will enable an even+even or an odd+odd operation which can provide 100 as a solution. Therefore we can say that more than 50 numbers on the blackboard without getting 100 in total is not possible.
– Prakhar
Slight error in explaining : he says 1 through 9 is not even in consideration 🙂