Bi 4. For any real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying the inequality $x_{1}+x_{2}+\cdots+x_{n} \leqslant \frac{1}{2}$. Find the minimum value of $\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n}\right)$.
When $x_{1}, x_{2}, \cdots, x_{n-2}, x_{n-1}+x_{n}$ are all constants, from the relation
$$
\begin{array}{l}
\left(1-x_{n-1}\right)\left(1-x_{n}\right) \\
=1-\left(x_{n-1}+x_{n}\right)+x_{n-1} x_{n}
\end{array}
$$

it is evident that the larger $\left|x_{n-1}-x_{n}\right|$, the smaller the value of the above expression. Let
$$
\begin{array}{l}
x_{i}{ }^{\prime}=x_{i}, i=1,2, \cdots, n-2, \\
x_{n-1}{ }^{\prime}=x_{n-1}+x_{n}, x_{n}{ }^{\prime}=0,
\end{array}
$$

then $x_{n-1}{ }^{\prime}+x_{n}{ }^{\prime}=x_{n-1}+x_{n}, x_{n-1}{ }^{\prime} x_{n}{ }^{\prime}=0 \leqslant$ $x_{n-1} x_{n}$. Therefore, we have
$$
\begin{array}{l}
\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n}\right) \\
\geqslant\left(1-x_{1}{ }^{\prime}\right)\left(1-x_{2}{ }^{\prime}\right) \cdots\left(1-x_{n-1}{ }^{\prime}\right),
\end{array}
$$

where $x_{1}{ }^{\prime}+x_{2}{ }^{\prime}+\cdots+x_{n-1}{ }^{\prime}=x_{1}+x_{2}+\cdots+x_{n} \leqslant \frac{1}{2}$. After at most $n-2$ such transformations, we obtain
$$
\begin{array}{l}
\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n}\right) \\
\geqslant 1-\left(x_{1}+x_{2}+\cdots+x_{n}\right) \geqslant \frac{1}{2},
\end{array}
$$

where equality holds when $x_{1}=\frac{1}{2}, x_{2}=x_{3}=\cdots=x_{n}=0$. Therefore, the minimum value sought is $\frac{1}{2}$.