11. Let $n$ positive numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=2023, M$ be
$$
\frac{a_{1}}{2023+a_{1}}, \frac{a_{2}}{2023+a_{1}+a_{2}}, \cdots, \frac{a_{n}}{2023+a_{1}+a_{2}+\cdots+a_{n}}
$$

the maximum value among these, then the minimum possible value of $M$ is $\qquad$ .
Let $x_{0}=2023, x_{1}=2023+a_{1}, x_{2}=2023+a_{1}+a_{2}, \cdots$,
$x_{n}=2023+a_{1}+a_{2}+\cdots+a_{n}=4096$, then $a_{k}=x_{k}-x_{k-1}(k=1,2, \cdots, n)$.
Thus, $M \geqslant \frac{a_{k}}{2023+a_{1}+a_{2}+\cdots+a_{k}}=\frac{x_{k}-x_{k-1}}{x_{k}}=1-\frac{x_{k-1}}{x_{k}} \Rightarrow 1-M \leqslant \frac{x_{k-1}}{x_{k}}$.
Therefore, $(1-M)^{n} \leqslant \prod_{k=1}^{n} \frac{x_{k-1}}{x_{k}}=\frac{x_{0}}{x_{n}}=\frac{2023}{4096}=\frac{1}{2} \Rightarrow 1-M \leqslant \frac{1}{\sqrt[n]{2}} \Rightarrow M \geqslant 1-\frac{1}{\sqrt[n]{2}}$,
Equality holds when $\frac{x_{k-1}}{x_{k}}=\sqrt[n]{2} \Rightarrow x_{k}=2023 \cdot(\sqrt[n]{2})^{k} \Rightarrow a_{k}=2023\left[(\sqrt[n]{2})^{k}-(\sqrt[n]{2})^{k-1}\right]$.
Therefore, the minimum possible value of $M$ is $1-\frac{1}{\sqrt[n]{2}}$.