Solution. Let

$$
M=\left\{x_{1}, \ldots, x_{n}\right\}, \quad x_{1}+\cdots+x_{n}=S
$$

Replace the element $x_{1}$ with the sum of the others. Then

$$
S=\left(S-x_{1}\right)+x_{2}+x_{3}+\cdots+x_{n}=\left(S-x_{1}\right)+\left(S-x_{1}\right)
$$

Reasoning similarly for the other elements, we get that

$$
2 x_{k}=S, \quad k=1 \ldots n \text {. }
$$

Thus, all elements of the set are equal to each other. Since the sum does not change when one addend is replaced, this addend must be equal to what it is replaced with, i.e.,

$$
x_{1}=x_{2}+x_{3}+\cdots+x_{n}
$$

Considering the equality of the elements, we get $x_{1}=(n-1) x_{1}$, hence, $x_{1}=0$. Therefore, the product of all numbers in the set $M$ is 0.

## Answer: 0.
Problem 3. The set $M$ consists of $n$ numbers, $n$ is odd, $n>1$. It is such that when any of its elements is replaced by the sum of the other $n-1$ elements from $M$, the sum of all $n$ elements does not change. Find the product of all $n$ elements of the set $M$.