17. We use the following obvious consequences of $(a+b)^{2} \geq 4 a b$ :
$$\begin{aligned} & \frac{1}{\left(a_{1}+a_{2}\right)\left(a_{3}+a_{4}\right)} \geq \frac{4}{\left(a_{1}+a_{2}+a_{3}+a_{4}\right)^{2}} \\ & \frac{1}{\left(a_{1}+a_{4}\right)\left(a_{2}+a_{3}\right)} \geq \frac{4}{\left(a_{1}+a_{2}+a_{3}+a_{4}\right)^{2}} \end{aligned}$$
Now we have
$$\begin{aligned} & \frac{a_{1}+a_{3}}{a_{1}+a_{2}}+\frac{a_{2}+a_{4}}{a_{2}+a_{3}}+\frac{a_{3}+a_{1}}{a_{3}+a_{4}}+\frac{a_{4}+a_{2}}{a_{4}+a_{1}} \\ = & \frac{\left(a_{1}+a_{3}\right)\left(a_{1}+a_{2}+a_{3}+a_{4}\right)}{\left(a_{1}+a_{2}\right)\left(a_{3}+a_{4}\right)}+\frac{\left(a_{2}+a_{4}\right)\left(a_{1}+a_{2}+a_{3}+a_{4}\right)}{\left(a_{1}+a_{4}\right)\left(a_{2}+a_{3}\right)} \\ \geq & \frac{4\left(a_{1}+a_{3}\right)}{a_{1}+a_{2}+a_{3}+a_{4}}+\frac{4\left(a_{2}+a_{4}\right)}{a_{1}+a_{2}+a_{3}+a_{4}}=4 . \end{aligned}$$