265 It is known that for all positive integers $n$,
$$
\prod_{i=1}^{n}\left(1+\frac{1}{3 i-1}\right) \geqslant \frac{k}{2} \sqrt[3]{19 n+8}
$$

always holds. Try to find the maximum value of $k$.
Let $T_{n}=\prod_{i=1}^{n}\left(1+\frac{1}{3 i-1}\right)$. Then
$$
\frac{T_{n+1}}{T_{n}}=1+\frac{1}{3 n+2}=\frac{3 n+3}{3 n+2} \text {. }
$$

Given $T_{n} \geqslant \frac{k}{2} \sqrt[3]{19 n+8}$, we have $k \leqslant \frac{2 T_{n}}{\sqrt[3]{19 n+8}}$.
Let $f(n)=\frac{2 T_{n}}{\sqrt[3]{19 n+8}}$. Then
$$
\begin{array}{l}
f(n+1)=\frac{2 T_{n+1}}{\sqrt[3]{19 n+27}}, \\
\frac{f(n+1)}{f(n)}=\frac{2 T_{n+1}}{\sqrt[3]{19 n+27}} \cdot \frac{\sqrt[3]{19 n+8}}{2 T_{n}} \\
=\frac{T_{n+1}}{T_{n}} \cdot \frac{\sqrt[3]{19 n+8}}{\sqrt[3]{19 n+27}} \\
=\frac{3 n+3}{3 n+2} \sqrt[3]{\frac{19 n+8}{19 n+27}} \\
=\sqrt[3]{\frac{(3 n+3)^{3}(19 n+8)}{(3 n+2)^{3}(19 n+27)}} .
\end{array}
$$

Notice that
$$
\begin{aligned}
& (3 n+3)^{3}(19 n+8)-(3 n+2)^{3}(19 n+27) \\
= & (19 n+8)\left[(3 n+3)^{3}-(3 n+2)^{3}\right]- \\
& 19(3 n+2)^{3} \\
= & (19 n+8)\left(27 n^{2}+45 n+19\right)- \\
& 19\left(27 n^{3}+54 n^{2}+36 n+8\right) \\
= & 45 n^{2}+37 n>0 .
\end{aligned}
$$

Then $(3 n+3)^{3}(19 n+8)$
$$
\begin{array}{l}
>(3 n+2)^{3}(19 n+27) . \\
\text { Hence } \sqrt[3]{\frac{(3 n+3)^{3}(19 n+8)}{(3 n+2)^{3}(19 n+27)}}>1 \\
\Rightarrow \frac{f(n+1)}{f(n)}>1 \Rightarrow f(n)<f(n+1) .
\end{array}
$$

Therefore, $f(n)$ is a monotonically increasing function with respect to $n$.
Since $T_{n} \geqslant \frac{k}{2} \sqrt[3]{19 n+8}$ holds for all positive integers $n$,
$$
\begin{array}{l}
k \leqslant \min \{f(n) \mid n=1,2, \cdots\}=f(1) \\
=\frac{2 T_{1}}{\sqrt[3]{19+8}}=\frac{2\left(1+\frac{1}{2}\right)}{\sqrt[3]{27}}=1 .
\end{array}
$$

Thus, the maximum value of $k$ is 1.