Find the maximum number of natural numbers $x_1,x_2, ... , x_m$  satisfying the conditions:
a) No $x_i - x_j , 1 \le i < j \le m$ is divisible by $11$,  and
b) The sum $x_2x_3 ...x_m + x_1x_3  ... x_m + \cdot \cdot \cdot + x_1x_2... x_{m-1}$ is divisible by $11$.
1. **Understanding the problem:**
   We need to find the maximum number of natural numbers \( x_1, x_2, \ldots, x_m \) such that:
   - No difference \( x_i - x_j \) (for \( 1 \leq i < j \leq m \)) is divisible by 11.
   - The sum \( x_2 x_3 \cdots x_m + x_1 x_3 \cdots x_m + \cdots + x_1 x_2 \cdots x_{m-1} \) is divisible by 11.

2. **Condition Analysis:**
   - For the first condition, if \( x_i \equiv x_j \pmod{11} \), then \( x_i - x_j \equiv 0 \pmod{11} \). Therefore, all \( x_i \) must have different remainders modulo 11.
   - For the second condition, we need to ensure that the sum of the products of all \( x_i \) except one is divisible by 11.

3. **Considering \( m = 11 \):**
   - If \( m = 11 \), then \( x_1, x_2, \ldots, x_{11} \) must be distinct modulo 11. This means they must be congruent to \( 0, 1, 2, \ldots, 10 \) modulo 11.
   - Without loss of generality, assume \( x_1 \equiv 0 \pmod{11} \). Then the product \( x_2 x_3 \cdots x_{11} \) must be divisible by 11.

4. **Using Wilson's Theorem:**
   - Wilson's theorem states that for a prime \( p \), \((p-1)! \equiv -1 \pmod{p}\).
   - For \( p = 11 \), we have \( 10! \equiv -1 \pmod{11} \). This implies \( 10! \) is not divisible by 11.

5. **Conclusion for \( m = 11 \):**
   - Since \( 10! \equiv -1 \pmod{11} \), it is not divisible by 11. Therefore, the condition \( x_2 x_3 \cdots x_{11} \equiv 0 \pmod{11} \) cannot be satisfied.
   - Hence, \( m = 11 \) is not possible.

6. **Considering \( m = 10 \):**
   - If \( m = 10 \), we can choose \( x_1, x_2, \ldots, x_{10} \) such that they are distinct modulo 11 and none of them is 0 modulo 11.
   - This ensures that no \( x_i - x_j \) is divisible by 11.
   - The sum \( x_2 x_3 \cdots x_{10} + x_1 x_3 \cdots x_{10} + \cdots + x_1 x_2 \cdots x_9 \) must be divisible by 11.

7. **Verification for \( m = 10 \):**
   - Since \( x_1, x_2, \ldots, x_{10} \) are distinct and non-zero modulo 11, their product modulo 11 is \( 1 \cdot 2 \cdot \ldots \cdot 10 \equiv -1 \pmod{11} \).
   - The sum of the products of all \( x_i \) except one will be divisible by 11 because each term will be a product of 9 distinct non-zero elements modulo 11, and the sum of such terms will be congruent to 0 modulo 11.

8. **Conclusion:**
   - The maximum number of natural numbers \( x_1, x_2, \ldots, x_m \) satisfying the given conditions is \( m = 10 \).

The final answer is \( \boxed{10} \)