61. Let $a_{i} \in \mathbf{R}, i=1,2,3,4,5$, find
$$\frac{a_{1}}{a_{2}+3 a_{3}+5 a_{4}+7 a_{5}}+\frac{a_{2}}{a_{3}+3 a_{4}+5 a_{5}+7 a_{1}}+\cdots+\frac{a_{5}}{a_{1}+3 a_{2}+5 a_{3}+7 a_{4}}$$

the minimum value. (2004 Jilin Province Mathematics Competition Problem)
61. Let the original expression be $A$, by Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
A\left(a_{1}\left(a_{2}+3 a_{3}+5 a_{4}+7 a_{5}\right)+a_{2}\left(a_{3}+3 a_{4}+5 a_{5}+7 a_{1}\right)+\cdots+\right. \\
\left.a_{5}\left(a_{1}+3 a_{2}+5 a_{3}+7 a_{4}\right)\right) \geqslant\left(a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\right)^{2}
\end{array}$$

Thus, we have
$$\frac{A \geqslant\left(a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\right)^{2}}{8 \sum_{1 \leqslant i<j \leqslant 5} a_{i} a_{j}}$$

Since
$$4\left(a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\right)^{2}-10 \sum_{1 \leqslant i \leqslant j \leqslant 5} a_{i} a_{j}=\sum_{1 \leqslant i<j \leqslant 5}\left(a_{i}-a_{j}\right)^{2} \geqslant 0$$

Therefore, $\left(a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\right)^{2} \geqslant \frac{5}{2} \sum_{1 \leqslant i<j \leqslant 5} a_{i} a_{j}$, hence $A \geqslant \frac{5}{16}$.
When $a_{1}=a_{2}=a_{3}=a_{4}=a_{5}$, the equality in (1) and (2) holds, i.e., $A=\frac{5}{56}$. In summary, the minimum value sought is $\frac{5}{16}$.