Find the maximum number of natural numbers $x_{1}, x_{2}, \ldots, x_{m}$ satisfying the conditions:

a) No $x_{i}-x_{j}, 1 \leq i<j \leq m$ is divisible by 11 ; and

b) The sum $x_{2} x_{3} \ldots x_{m}+x_{1} x_{3} \ldots x_{m}+\cdots+x_{1} x_{2} \ldots x_{m-1}$ is divisible by 11.
The required maximum is 10.

According to a), the numbers $x_{i}, 1 \leq i \leq m$, are all different $(\bmod 11)$ (1)

Hence, the number of natural numbers satisfying the conditions is at most 11.

If $x_{j} \equiv 0(\bmod 11)$ for some $j$, then

$$
x_{2} x_{3} \ldots x_{m}+x_{1} x_{3} \ldots x_{m}+\cdots+x_{1} x_{2} \ldots x_{m-1} \equiv x_{1} \ldots x_{j-1} x_{j+1} \ldots x_{m} \quad(\bmod 11)
$$

which would lead to $x_{i} \equiv 0(\bmod 11)$ for some $i \neq j$, contradicting (1).

We now prove that 10 is indeed the required maximum.

Consider $x_{i}=i$, for all $i \in\{1,2, \ldots, 10\}$. The products $2 \cdot 3 \cdots \cdot 10,1 \cdot 3 \cdots \cdots 10, \ldots$, $1 \cdot 2 \cdots \cdot 9$ are all different $(\bmod 11)$, and so

$$
2 \cdot 3 \cdots \cdots 10+1 \cdot 3 \cdots \cdots 10+\cdots+1 \cdot 2 \cdots \cdot 9 \equiv 1+2+\cdots+10 \quad(\bmod 11)
$$

and condition b) is satisfied, since $1+2+\cdots+10=55=5 \cdot 11$.