6. $100 n$ positive numbers $x_{1}, x_{2}, \cdots, x_{n}$ sum up to 1. Let $S$ be the largest of the following numbers:
$$\frac{x_{1}}{1+x_{1}}, \frac{x_{2}}{1+x_{1}+x_{2}}, \cdots, \frac{x_{n}}{1+x_{1}+x_{2}+\cdots+x_{n}} .$$

Find the smallest possible value of $S$. For what values of $x_{1}, x_{2}, \cdots, x_{n}$ is this minimum value achieved?
[Solution] Let $y_{0}=1, y_{k}=1+x_{1}+x_{2}+\cdots+x_{k}(1 \leqslant k \leqslant n)$.
Then $\square$
$$x_{k}=y_{k}-y_{k-1}$$

From the given information,
$$\frac{x_{k}}{1+x_{1}+x_{2}+\cdots+x_{k}} \leqslant S$$

which is $\frac{y_{k}-y_{k-1}}{y_{k}} \leqslant S$,
thus we have $1-S \leqslant \frac{y_{k-1}}{y_{k}} \quad(k=1,2, \cdots, n)$
From the given information, we know that $S \leqslant 1$ and $y_{k}>0(k=0,1,2, \cdots, n)$, so we can multiply the $n$ inequalities expressed by (1), thus obtaining
$$(1-S)^{n} \leqslant \frac{y_{0}}{y_{n}}$$

Noting that $y_{0}=1, y_{n}=2$, we have
$$\begin{array}{l}
(1-S)^{n} \leqslant \frac{1}{2} \\
S \geqslant 1-2^{-\frac{1}{n}}
\end{array}$$

On the other hand, if we take
$$\frac{y_{k-1}}{y_{k}}=1-S=2^{-\frac{1}{n}} \quad(k=1,2, \cdots, n)$$

that is, taking $\square$
$$y_{0}=1, y_{1}=2^{\frac{1}{n}}, y_{2}=2^{\frac{2}{n}}, y_{3}=2^{\frac{3}{n}}, \cdots, y_{n}=2$$

then $\square$
$$S=1-2^{-\frac{1}{n}}$$

In this case, $x_{k}=y_{k}-y_{k-1}=2^{\frac{k}{n}}-2^{\frac{k-1}{n}} \quad(k=1,2, \cdots, n)$.
This means that when $x_{k}=2^{\frac{k}{n}}-2^{\frac{k-1}{n}}(k=1,2, \cdots, n)$, $S$ can achieve the minimum value $1-2^{-\frac{1}{n}}$.