Bump, any idea $?$

How can there be distinct numbers when all the numbers are nonnegative and less than 1, implying that they are zero?

We are dealing with all reals, not just integers I think.

Put the numbers in increasing order. At most $a_1=0$. Clearly for $N=9$ we can find a solution. Now assume that $N$ is even $\ge 10$. Without loss of generality we can assume that $a_1+\cdots +a_9=n_1$, where $n_1$ is an integer. Let $S= a_1+\cdots +a_7$. Then $S+a_{10}+a_i$ can be an integer only if $i\ge11$ (because all the numbers are distinct and smaller than 1). That is, the case $N=10$ is not possible. So assume $N\ge 12$ and let $S+a_{10}+a_{11}=n_2$, where $n_2$ is an integer. Now look at $S+a_{12}+a_i$; this can be an integer only if $i\ge 13$. So the case $N=12$ is not possible. And so on for even $N$. Now assume that $N$ is odd and start with $N=11$. As before, WLOG we can assume that $a_1+\cdots +a_9=n_1$, and $a_1+\cdots +a_7+a_{10}+a_{11}=n_2$, where $n_1$ and $n_2$ are integers. Clearly we must have $n_2-n_1=1$ or $(a_{10}+a_{11})-(a_8+a_9)=1$ because all the numbers are distinct, positive (except maybe $a_1=0$) and smaller than 1. Now let $S= a_1+\cdots +a_6$. Then $S+a_9+a_{10}+a_i$ can be an integer only if $i\ge12$. That is, the case $N=11$ is not possible. So assume $N\ge 13$ and let $S+a_9+a_{10}+a_{12}=n_3$, where $n_3$ is an integer. But now $n_3-n_2=1$ and $n_3-n_1=1$, which leads to contradiction. And so on... The only solution is $N=9$.

<tk1> wrote: Now assume that $N$ is even $\ge 10$. Without loss of generality we can assume that $a_1+\cdots +a_9=n_1$, where $n_1$ is an integer. Well.. It DOES LOSE generality... Because you already put the $a_i$ in order you can no more assumpt "without loss of genarality"

Does anyone have an official solution for this?

Official solution: For $N=9$ obvious. Let $N>9$, and numbers are $a_1,...,a_N$ $S=a_1+...+a_N$ and $T=a_1+...+a_7$ First, prove that for every $i>7$ exist only one index $j>7$ such that $T+a_i+a_j$ is integer. Let for some $i$ there are two such indexes $j$ and $k$. Then $1>|a_k-a_j|=|T+a_i+a_j-(T+a_i+a_k)|=|integer - integer|>0$ - contradiction. We call pair $(i,j)$ as fitting if $T+a_i+a_j$ is integer. So all indexes $>7$ can be divided in fitting pairs $(x_1,y_1),...(x_m,y_m)$ where $x_i$ and $y_i$ is $a_j$ for some $7<j<N$. So $N=7+2m$. Build sum as $P=\sum_{i=1}^{m}{(T+x_i+y_i)}=T*(m-1)+S$. Sum $P$ is integer. And $T=\frac{P-S}{m-1}$ Similar way we set $T`=a_2+...+a_8$ and $T`=\frac{P`-S}{m-1}$ $a_8-a_1=T`-T=\frac{P`-P}{m-1}$ Let $a_1$ is minimal. So it follows, that for $1<i<N$ we can write $a_i=a_1+\frac{b_i}{m-1}$ where $b_i$ - positive integer , and $N=7+2m$ $1>a_N\geq a_1+\frac{N-1}{m-1} \to N<m$ but $N=7+2m>m$ contradiction.

Thanks!! If you know, what is the motive for thinking $T=a_1+...+a_7$