Adding $4$ to both sides of the second equation, we write:
$$1 + a + \frac{1+a+b}{1+a} + \frac{1+a+b+c}{1+a+b} + \frac{1+a+b+c+d}{1+a+b+c} = 12.$$
Applying the AM-GM inequality successively, we obtain:
$$\begin{aligned} 1 + a + \frac{1+a+b}{1+a} &\ge 2\sqrt{(1+a) \cdot \frac{1+a+b}{1+a}} = 2\sqrt{1+a+b}; \\ \frac{1+a+b+c}{1+a+b} + \frac{1+a+b+c+d}{1+a+b+c} &\ge 2\sqrt{\frac{1+a+b+c}{1+a+b} \cdot \frac{1+a+b+c+d}{1+a+b+c}} = \\ &\ge 2\sqrt{\frac{81}{1+a+b}}. \end{aligned}$$
By adding these two inequalities and applying the AM-GM inequality again, we have:
$$\begin{aligned} 12 &= 1 + a + \frac{1+a+b}{1+a} + \frac{1+a+b+c}{1+a+b} + \frac{1+a+b+c+d}{1+a+b+c} \\ &\ge 2\sqrt{1+a+b} + \frac{18}{\sqrt{1+a+b}} \ge 12. \end{aligned}$$

From this, we obtain $a + b = 8$ and
$$1 + a = \frac{1 + a + b}{1 + a} = \frac{1 + a + b + c}{1 + a + b} = \frac{1 + a + b + c + d}{1 + a + b + c} = 3.$$
Therefore, the desired numbers are $a = 2$, $b = 6$, $c = 18$, $d = 54$.