Let $\Phi$.

On the board, $N \geq 9$ different non-negative numbers, each less than one, are written. It turns out that for any eight different numbers on the board, there is a ninth, different from them, such that the sum of these nine numbers is an integer. For which $N$ is this possible?
Clearly, when $N=9$, the required is possible - it is sufficient to write 9 different positive numbers with a unit sum on the board. We will show that when $N>9$, the required is impossible.

Assume the contrary; let $S$ be the sum of all numbers on the board. Choose any numbers $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{7}$ with sum $T$; let $A$ be the set of all other numbers on the board. By the condition, for any number $\beta \in A$, there exists a number $\gamma \in A$ different from it such that the number $T+\beta+\gamma$ is an integer. We say that the number $\gamma$ corresponds to the number $\beta$. Note that such a number is unique. Indeed, if there were another number $\gamma^{\prime} \in A$ for which the sum $T+\beta+\gamma^{\prime}$ is an integer, then the number $\gamma-\gamma^{\prime}=(T+\beta+\gamma)-\left(T+\beta+\gamma^{\prime}\right)$ would also be an integer. But this is impossible, since $0 < \gamma - \gamma^{\prime} < 1$, as $N=7+2 l>9$.

Consider the sum $\Sigma=\left(T+\beta_{1}+\gamma_{1}\right)+\left(T+\beta_{2}+\gamma_{2}\right)+\ldots+\left(T+\beta_{l}+\gamma_{l}\right)$. $\Sigma$ is an integer. On the other hand, each number from $A$ appears in $\Sigma$ exactly once; hence, $\Sigma=I T+(S-T)=S+(l-1) T$, from which $T=\frac{\Sigma-S}{\ell-1}$.

Now, choosing the numbers $\alpha_{2}, \alpha_{3}, \ldots, \alpha_{8}$ on the board and denoting their sum by $T^{\prime}$, we similarly obtain that $T^{\prime}=\frac{\Sigma^{\prime}-S}{\ell-1}$ for an integer $\Sigma^{\prime}$. Therefore, $\alpha_{1}-\alpha_{8}=\frac{\Sigma-S}{\ell-1}-\frac{\Sigma^{\prime}-S}{\ell-1}=\frac{\Sigma-\Sigma^{\prime}}{\ell-1}$.

Since $\alpha_{1}$ and $\alpha_{8}$ could be any two numbers on the board, we get that the difference between any two numbers on the board is of the form $\frac{k}{l-1}$ for an integer $k$.

Let $\mu$ be the smallest number on the board. Then the board can only contain the numbers $\mu, \mu+\frac{1}{\ell-1}, \ldots, \mu+\frac{\ell-2}{\ell-1}$ (all larger numbers would already be at least 1) - a total of $l$ numbers. However, the total number of numbers on the board is $N=7+2 l>l$; hence, they cannot all be distinct. Contradiction.

## Answer

Only when $N=9$.