[Proof] By the given condition, $x_{1}, x_{2}, x_{3}, x_{4}$ are four distinct roots of $P(x)-2$, therefore,
$$P(x)-2=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right) Q(x) .$$

where $Q(x)$ is a polynomial with integer coefficients or an integer.
No matter what integer $x$ is, $x-x_{1}, x-x_{2}, x-x_{3}, x-x_{4}$, and $Q(x)$ are all integers, and since $x_{1}, x_{2}, x_{3}, x_{4}$ are four different integers, $x-x_{1}, x-x_{2}, x-x_{3}, x-x_{4}$ are also not equal.
If $P(x)$ equals any one of $1,3,5,7,9$, then
$$P(x)-2=-1,1,3,5,7 .$$

any one of $-1, 1, 3, 5, 7$.
But none of $-1, 1, 3, 5, 7$ can be factored into the product of four distinct factors. Therefore, $P(x)$ cannot equal any one of $1,3,5,7,9$.