8. On the blackboard, there are $N(N \geqslant 9)$ distinct non-negative real numbers less than 1. It is known that for any eight numbers on the blackboard, there exists another number on the blackboard such that the sum of these nine numbers is an integer. Find all possible values of $N$.

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8. $N=9$.
An example for $N=9$ exists. It is sufficient for the sum of these nine numbers to be an integer.
Now, let $N>9$. Let $S$ denote the sum of all numbers.
Take any seven numbers $a_{1}, a_{2}, \cdots, a_{7}$, and let $T$ denote their sum, and $A$ denote the complement of the set formed by these seven numbers.

Take any $c \in A$, there exists $d \in A$ such that $T+c+d$ is an integer.
It is easy to see that such a $d$ is unique.
In fact, if $T+c+d$ and $T+c+d^{\prime}$ are both integers, then
$$
\left|d-d^{\prime}\right|=\left|(T+c+d)-\left(T+c+d^{\prime}\right)\right|
$$

is an integer.
Thus, $d=d^{\prime}$.
The above $d$ is paired with $c$. Thus, the numbers in set $A$ are paired as follows:
$$
\left(c_{1}, d_{1}\right),\left(c_{2}, d_{2}\right), \cdots,\left(c_{l}, d_{l}\right),
$$

where $l=\frac{N-7}{2}>1$.
Consider the sum
$$
S_{1}=\sum_{i=1}^{l}\left(T+c_{i}+d_{i}\right)=l T+(S-T)
$$

which is an integer.
Thus, $T=\frac{S_{1}-S}{l-1}$.
Now, take the numbers $a_{2}, a_{3}, \cdots, a_{8}$ on the board. Let their sum be $T^{\prime \prime}$.
As before, we get $T^{\prime}=\frac{S_{1}^{\prime}-S}{l-1}$.
Therefore, $a_{1}-a_{8}=\frac{S_{1}-S}{l-1}-\frac{S_{1}^{\prime}-S}{l-1}=\frac{S_{1}-S_{1}^{\prime}}{l-1}$.
Since $a_{1}$ and $a_{8}$ can be any two numbers on the board, we conclude that the difference between any two numbers on the board has the form $\frac{k}{l-1}(k \in \mathbf{Z})$.

Let $a$ be the smallest number on the board. Then the numbers on the board belong to the set
$$
\left\{a, a+\frac{1}{l-1}, a+\frac{2}{l-1}, \cdots, a+\frac{l-2}{l-1}\right\}
$$

which contains at most $l-1$ numbers.
This contradicts the fact that the number of numbers is $N=7+2 l>l$.