Solution:

The smallest possible value for $S$ is $8$.
To show that $S \geq 8$ always holds, we apply the AM-GM inequality twice:

$\left(\frac{a}{b}+\frac{c}{d}\right)+\left(\frac{b}{c}+\frac{d}{a}\right) \geq 2 \sqrt{\frac{a c}{b d}}+2 \sqrt{\frac{b d}{a c}}=\frac{2(a c+b d)}{\sqrt{a b c d}}=\frac{2(a+c)(b+d)}{\sqrt{a b c d}} \geq 2 \cdot \frac{2 \sqrt{a c} \cdot 2 \sqrt{b d}}{\sqrt{a b c d}}=8$.

Equality holds at these inequalities when $a=c$ and $b=d$. Thus $(a+c)(b+d)=a c+b d$ can be rewritten as $4 a b=a^{2}+b^{2}$. Solving for $a$ gives $a=b(2 \pm \sqrt{3})$ and thus a possible solution $S=8$ for $b=d=1$ and $a=c=2+\sqrt{3}$.

Solution:

The shift $(a, b, c, d) \rightarrow (b, c, d, a)$ changes neither the value of $S$ nor the constraint. Therefore we may assume WLOG that $a c \geq b d$ and set $t=\sqrt{\frac{a c}{b d}}$ with $t \geq 1$.

From the condition it follows that $t^{2}+1=\frac{a c+b d}{b d}=(a+c)\left(\frac{1}{d}+\frac{1}{b}\right) \geq 2 \sqrt{a c} \cdot 2 \sqrt{\frac{1}{b d}}=4 t$, where in the second-to-last step the AM-GM inequality was used. Thus $t^{2}-4 t+1 \geq 0$ holds and therefore $(t \geq 2+\sqrt{3} \vee t \leq 2-\sqrt{3})$, which, because of $t \geq 1$, leads to $t \geq 2+\sqrt{3}$.

We have $S=\left(\frac{a}{b}+\frac{c}{d}\right)+\left(\frac{b}{c}+\frac{d}{a}\right) \geq 2 \sqrt{\frac{a c}{b d}}+2 \sqrt{\frac{b d}{a c}}=2\left(t+\frac{1}{t}\right)$. By differentiating we see that for all $t \geq 2+\sqrt{3}$ we have: $S'(t)=2-2 / t^{2} \geq 0$. Therefore $S$ has a minimum at $t=2+\sqrt{3}$ and it follows that $S \geq 2\left(t+\frac{1}{t}\right) \geq 2\left(2+\sqrt{3}+\frac{1}{2+\sqrt{3}}\right)=2(2+\sqrt{3}+2-\sqrt{3})=8$.

For $(a, b, c, d)=(1,2+\sqrt{3}, 1,2+\sqrt{3})$ both $(a+c)(b+d)=8+4 \sqrt{3}=1 \cdot 1+(2+\sqrt{3})(2+\sqrt{3})=a c+b d$ and $S=\frac{1}{2+\sqrt{3}}+\frac{2+\sqrt{3}}{1}+\frac{1}{2+\sqrt{3}}+\frac{2+\sqrt{3}}{1}=2-\sqrt{3}+2+\sqrt{3}+2-\sqrt{3}+2+\sqrt{3}=8$ hold.

Therefore $8$ is the smallest value of $S$. $\square$