## Solution.

Let $\overline{a b c}$ be one such number.

Then the numbers $\overline{a c b}, \overline{b a c}, \overline{b c a}, \overline{c a b}, \overline{c b a}$ are among the numbers we are adding.

Their sum is

$$\begin{aligned} \overline{a b c}+\overline{a c b}+\overline{b a c}+\overline{b c a}+\overline{c a b}+\overline{c b a} & =(100 a+10 b+c)+\ldots+(100 c+10 b+a) \\ & =222 a+222 b+222 c=222(a+b+c) . \end{aligned}$$

Therefore, the sum of all such numbers is divisible by 222, 1 point

which means it is divisible by 2, 3, and 37.