Combinatorics: numbers on a blackboard with integer sums

$N\geq9$ distinct real numbers are written on a blackboard. All these numbers are nonnegative, and all are less than $1$. It happens that for every $8$ distinct numbers on the board, the board contains a ninth number distinct from the eight such that the sum of all these nine numbers is integer.

Find all values $N$ for which this is possible.

It's trivial to check N=9 works, and N=10,11 don't work. Not sure after that...

Edit: To see why it doesn't work for N=10:

Let the numbers be $a_1,..a_{10}$ and their sum be $S$. By considering the first $8$ numbers, we find that at least one of $S-a_9$ or $S-a_{10}$ is a positive integer. By considering the last $8$ numbers, we get one of $S-a_1$ and $S-a_2$ is a positive integer. WLOG $S-a_1$ and $S-a_9$ are both positive integers. Their difference is $a_9-a_1$, which must be an integer. By $a_1$ and $a_9$ are nonnegative and less than one, so this is impossible.

A similar argument works for $N=11$, but it sort of breaks down for $N=12$.

• I don't see why it doesn't work for any N. Perhaps you should explain why it doesn't for N = 10. – Doug M May 27 '16 at 23:37
• OK see my edit. – Joshua Benabou May 28 '16 at 0:12
• 1) What is the source of this problem, please? 2) Can you solve the problem where instead of 8 distinct numbers you're interested in 2 distinct numbers? or 3? Maybe the small cases will give you a clue as to how to deal with the big one. – Gerry Myerson May 28 '16 at 1:31
• It's from a Russian math contest, but there aren't any solutions available online as far as I know. As far as smaller cases, I'll try that. – Joshua Benabou May 28 '16 at 2:52

This is a partial answer, which shows that it's not possible for even $N$.
Assume that it is possible for some even number $N = 2m\geq10$. Without loss of generality we can assume that $$a_1+...+a_9 = n_1,$$ where $n_1$ is an integer. Consider the sum of $a_1,...,a_7$ and one of the other numbers, which we call $a_{10}$. Since $a_1+...+a_9$ is an integer and since all the numbers are distinct and smaller than 1, it follows that $a_1+...+a_7+a_{10}+a_i$, where $i=8$ or $i=9$ cannot be an integer. If $N = 10$, this means that there is no number $a_j$ such that $a_1+...+a_7+a_{10}+a_j$ is an integer, and we have a contradiction. If $N=2m>10$, we can find such a number. Let's call this number $a_{11}$, so $$a_1+...+a_7 + a_{10} + a_{11} = n_2,$$ where $n_2$ is an integer. Once more, consider the sum of $a_1,...,a_7$ and one number not already chosen, which we call $a_{12}$. Again, since the numbers are distinct it must follow that the sum $a_1+...+a_7+a_{12}+a_i$ cannot be an integer if $i = 8,9,10,11$. Continuing this line of argument, we arrive at the fact that $a_1+...+a_7+a_N+a_i$ cannot be an integer if $i = 8,9,...,N-1$, but since we have no numbers left we arrive at a contradiction, and thus it cannot be done for even numbers.