$N\geq9$ distinct real numbers are written on a blackboard. All these numbers are nonnegative, and all are less than $1$. It happens that for very $8$ distinct numbers on the board, the board contains the ninth number distinct from eight such that the sum of all these nine numbers is integer. Find all values $N$ for which this is possible. [i](F. Nilov)[/i]
1. **Initial Setup and Assumptions:**
   - We are given \( N \geq 9 \) distinct real numbers, all nonnegative and less than 1.
   - For any 8 distinct numbers, there exists a ninth number such that the sum of these nine numbers is an integer.
   - We need to find all possible values of \( N \).

2. **Case \( N = 9 \):**
   - Consider the numbers \( a_1, a_2, \ldots, a_9 \).
   - Since the sum of any 9 numbers is an integer, we can choose \( a_1, a_2, \ldots, a_9 \) such that \( a_1 + a_2 + \cdots + a_9 = n_1 \), where \( n_1 \) is an integer.
   - This satisfies the condition, so \( N = 9 \) is a possible solution.

3. **Case \( N \geq 10 \):**
   - Assume \( N \) is even and \( N \geq 10 \).
   - Without loss of generality, let \( a_1 + a_2 + \cdots + a_9 = n_1 \), where \( n_1 \) is an integer.
   - Let \( S = a_1 + a_2 + \cdots + a_7 \).
   - For \( S + a_{10} + a_i \) to be an integer, \( a_i \) must be distinct and less than 1. This implies \( i \geq 11 \).
   - Therefore, \( N = 10 \) is not possible.

4. **Generalizing for Even \( N \geq 12 \):**
   - Assume \( N \geq 12 \).
   - Let \( S + a_{10} + a_{11} = n_2 \), where \( n_2 \) is an integer.
   - For \( S + a_{12} + a_i \) to be an integer, \( a_i \) must be distinct and less than 1. This implies \( i \geq 13 \).
   - Therefore, \( N = 12 \) is not possible.
   - This pattern continues for all even \( N \).

5. **Case \( N \) is Odd and \( N \geq 11 \):**
   - Assume \( N = 11 \).
   - Let \( a_1 + a_2 + \cdots + a_9 = n_1 \) and \( a_1 + a_2 + \cdots + a_7 + a_{10} + a_{11} = n_2 \), where \( n_1 \) and \( n_2 \) are integers.
   - We must have \( n_2 - n_1 = 1 \) or \( (a_{10} + a_{11}) - (a_8 + a_9) = 1 \).
   - Let \( S = a_1 + a_2 + \cdots + a_6 \).
   - For \( S + a_9 + a_{10} + a_i \) to be an integer, \( a_i \) must be distinct and less than 1. This implies \( i \geq 12 \).
   - Therefore, \( N = 11 \) is not possible.

6. **Generalizing for Odd \( N \geq 13 \):**
   - Assume \( N \geq 13 \).
   - Let \( S + a_9 + a_{10} + a_{12} = n_3 \), where \( n_3 \) is an integer.
   - We must have \( n_3 - n_2 = 1 \) and \( n_3 - n_1 = 1 \), leading to a contradiction.
   - This pattern continues for all odd \( N \).

Conclusion:
- The only possible value for \( N \) is \( N = 9 \).

The final answer is \( \boxed{ N = 9 } \).