Solution:

Answer: $3$. To solve this problem, two parts are required. Part A shows that $L = 3$ works. Part B shows that no $L' < 3$ works.

Part A

We show that for all quadruples $(a, b, c, d)$ (with $1 < a < b < c < d$) we have
$$\frac{1}{a} + \frac{a}{b} + \frac{b}{c} + \frac{c}{d} \leqslant 3.$$
We do this by considering the two cases $b = a + 1$ and $b \geq a + 2$ separately.

If $b \geq a + 2$ then we can get
$$\frac{1}{a} + \frac{a}{b} + \frac{b}{c} + \frac{c}{d} \leqslant \frac{1}{a} + \frac{a}{a + 2} + \frac{b}{c} + \frac{c}{d}$$
$$\qquad < \frac{1}{a} + \frac{a}{a + 2} + 1 + 1$$
$$\qquad = \frac{2 - a}{a(a + 2)} + 3 \leqslant 3$$
The numerator $(2 - a)$ is non-positive because $1 < a$ and $a$ is an integer.

If $b = a + 1$ then we can get
$$\frac{1}{a} + \frac{a}{b} + \frac{b}{c} + \frac{c}{d} = \frac{1}{a} + \frac{a}{a + 1} + \frac{a + 1}{c} + \frac{c}{d}$$
$$\qquad < \frac{1}{a} + \frac{a}{a + 1} + \frac{a + 1}{c} + 1$$
$$\qquad \leqslant \frac{1}{a} + \frac{a}{a + 1} + \frac{a + 1}{a + 2} + 1$$
$$\qquad = \frac{2 - a^2}{a(a + 1)(a + 2)} + 3 < 3.$$
The numerator $(2 - a^2)$ is negative because $1 < a$ and $a$ is an integer.

In either case we get $\frac{1}{a} + \frac{a}{b} + \frac{b}{c} + \frac{c}{d} \leqslant 3$ for all quadruples $(a, b, c, d)$.

Part B

For the sake of contradiction, suppose there existed some $L' < 3$ which worked. We will show that there exists a quadruple $(a, b, c, d)$ such that
$$\frac{1}{a} + \frac{a}{b} + \frac{b}{c} + \frac{c}{d} > L'.$$
We will do this explicitly by choosing $a, b, c, d$ to be large consecutive integers. If $L' = 3 - \epsilon$ then we have $\epsilon > 0$, and so there exists some integer $n$ such that $n > \frac{3}{\epsilon}$. This ensures $\frac{1}{n + 1}$, $\frac{1}{n + 2}$ and $\frac{1}{n + 3}$ are each smaller than $\frac{\epsilon}{3}$.

Now consider $(a, b, c, d) = (n, n + 1, n + 2, n + 3)$.
$$\frac{1}{a} + \frac{a}{b} + \frac{b}{c} + \frac{c}{d} > \frac{a}{b} + \frac{b}{c} + \frac{c}{d}$$
$$\qquad = \frac{n}{n + 1} + \frac{n + 1}{n + 2} + \frac{n + 2}{n + 3}$$
$$\qquad = 3 - \left(\frac{1}{n + 1} + \frac{1}{n + 2} + \frac{1}{n + 3}\right)$$
$$\qquad > 3 - \left(\frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3}\right)$$
$$\qquad = L'$$
Contradiction.