Six, find the largest constant $M(M>0)$, such that for any positive integer $n$, there exist positive real number sequences $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$ satisfying
$$
\begin{array}{l}
\text { (1) } \sum_{k=1}^{n} b_{k}=1, \\
2 b_{k} \geqslant b_{k-1}+b_{k+1}(k=2,3, \cdots, n-1) ; \\
\text { (2) } a_{k}^{2} \leqslant 1+\sum_{i=1}^{k} a_{i} b_{i}\left(k=1,2, \cdots, n, a_{n}=M\right) .
\end{array}
$$
Lemma:
$$
\max _{1 \leqslant k \leqslant n}\left\{a_{k}\right\}\frac{n-k}{n-m} b_{m}, & m\frac{m}{2} b_{m}+\frac{n-m-1}{2} b_{m}=\frac{n-1}{2} b_{m} \text {. }
$$

This implies $b_{m}<\frac{2}{n-1}$. The lemma is proved.
Returning to the original problem.
Let $f_{0}=1$,
$$
f_{k}=1+\sum_{i=1}^{k} a_{i} b_{i}(k=1,2, \cdots, n) .
$$

Then $f_{k}-f_{k-1}=a_{k} b_{k} \leqslant b_{k} \sqrt{f_{k}}$.
From this, we get
$$
\begin{array}{l}
\sqrt{f_{k}}-\sqrt{f_{k-1}} \leqslant b_{k} \cdot \frac{\sqrt{f_{k}}}{\sqrt{f_{k}}+\sqrt{f_{k-1}}} \\
=b_{k}\left[\frac{1}{2}+\frac{f_{k}-f_{k-1}}{2\left(\sqrt{f_{k}}+\sqrt{f_{k-1}}\right)^{2}}\right] \\
<b_{k}\left[\frac{1}{2}+\frac{2 b_{k}}{2\left(\sqrt{f_{k}}+\sqrt{f_{k-1}}\right)^{2}}\right] \\
<b_{k}\left[\frac{1}{2}+\frac{b_{k}}{4}\right] \\
<b_{k}\left[\frac{1}{2}+\frac{1}{2(n-1)}\right] .
\end{array}
$$

Summing over $k$ from 1 to $n$ gives
$$
\begin{array}{l}
M=a_{n} \leqslant \sqrt{f_{n}} \\
<\sqrt{f_{0}}+\sum_{k=1}^{n} b_{k} \cdot\left[\frac{1}{2}+\frac{1}{2(n-1)}\right] \\
=\frac{3}{2}+\frac{1}{2(n-1)} .
\end{array}
$$

By the arbitrariness of $n$, we get $M_{\max } \leqslant \frac{3}{2}$.
$M=\frac{3}{2}$ is achievable. An example is as follows:
$$
a_{k}=1+\frac{k}{2 n}, b_{k}=\frac{1}{n}(k=1,2, \cdots, n),
$$

Then $a_{k}^{2}=\left(1+\frac{k}{2 n}\right)^{2} \leqslant 1+\sum_{i=1}^{k} \frac{1}{n}\left(1+\frac{i}{2 n}\right)$ holds.
In summary, the maximum value of $M$ is $\frac{3}{2}$.