Four. (20 points) Let $a_{1} \in \mathbf{R}_{+}, i=1,2, \cdots, 5$. Find
$$
\begin{array}{l}
\frac{a_{1}}{a_{2}+3 a_{3}+5 a_{4}+7 a_{5}}+\frac{a_{2}}{a_{3}+3 a_{4}+5 a_{5}+7 a_{1}}+ \\
\cdots+\frac{a_{5}}{a_{1}+3 a_{2}+5 a_{3}+7 a_{4}}
\end{array}
$$

the minimum value.
Let the original expression be $A$. By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
A \cdot\left[a_{1}\left(a_{2}+3 a_{3}+5 a_{4}+7 a_{5}\right)+a_{2}\left(a_{3}+3 a_{4}+\right.\right. \\
\left.\left.\quad 5 a_{5}+7 a_{1}\right)+\cdots+a_{5}\left(a_{1}+3 a_{2}+5 a_{3}+7 a_{4}\right)\right] \\
\quad \geqslant\left(a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\right)^{2} .
\end{array}
$$

Thus, we have $A \geqslant \frac{\left(\sum_{i=1}^{5} a_{i}\right)^{2}}{8 \sum_{1 \leqslant i<j<5} a_{i} a_{j}}$.
Since $4\left(\sum_{i=1}^{5} a_{i}\right)^{2}-10 \sum_{1 \leqslant i<j<5} a_{i} a_{j}$
$$
=\sum_{1 \leqslant i<j \leqslant 5}\left(a_{i}-a_{j}\right)^{2} \geqslant 0 \text {, }
$$

we have $\left(\sum_{i=1}^{5} a_{i}\right)^{2} \geqslant \frac{5}{2} \sum_{1<i<j<5} a_{i} a_{j}$.
Therefore, $A \geqslant \frac{5}{16}$.
When $a_{1}=a_{2}=a_{3}=a_{4}=a_{5}$, the equality in equations (1) and (2) holds, i.e., $A=\frac{5}{16}$.
In conclusion, the minimum value sought is $\frac{5}{16}$.