Solution:

Denote by $S_{k}^{(m)}$ the money of the $k$-th player, $k=1,2,3,4$, after the move and payment of the $m$-th, $m=1,2,3,4$, and $S_{k}^{(0)}=S$ in the beginning, $k=1,2,3,4$. Denote by $a_{i}$ the sum of points on the dices thrown by $A_{i}$. It follows from the game rules that $S_{k}^{(m)}=S_{k}^{(m-1)}+\frac{1}{a_{m}} S_{k}^{(m-1)}=S_{k}^{(m-1)} \frac{1+a_{m}}{a_{m}}$ for $k \neq m$ (i.e. when $A_{k}$ gets money) and
$$\begin{aligned} S_{k}^{(k)} & =S_{k}^{(k-1)}-\frac{1}{a_{k}} \sum_{i \neq k} S_{i}^{(k-1)}=S_{k}^{(k-1)}-\frac{1}{a_{k}}\left(4 S-S_{k}^{(k-1)}\right) \\ & =S_{k}^{(k-1)} \frac{1+a_{k}}{a_{k}}-\frac{4 S}{a_{k}} \end{aligned}$$
when $A_{k}$ pays.

Using these formulas we find the money of the four players in the end of the game. We obtain that
$$\begin{aligned} \frac{6 S}{5} & =S_{1}^{(4)}=P S-\frac{4 S\left(1+a_{2}\right)\left(1+a_{3}\right)\left(1+a_{4}\right)}{a_{1} a_{2} a_{3} a_{4}} \\ \frac{6 S}{5} & =S_{2}^{(4)}=P S-\frac{4 S\left(1+a_{3}\right)\left(1+a_{4}\right)}{a_{2} a_{3} a_{4}} \\ \frac{4 S}{5} & =S_{3}^{(4)}=P S-\frac{4 S\left(1+a_{4}\right)}{a_{3} a_{4}} \\ \frac{4 S}{5} & =S_{4}^{(4)}=P S-\frac{4 S}{a_{4}} \end{aligned}$$
where
$$P=\frac{\left(1+a_{1}\right)\left(1+a_{2}\right)\left(1+a_{3}\right)\left(1+a_{4}\right)}{a_{1} a_{2} a_{3} a_{4}}$$
By the first two equations we get $a_{2}=a_{1}-1$, and by the last two we have $a_{4}=a_{3}-1$. Now by the second and third equations we obtain
$$\frac{2}{5}=\frac{4}{a_{4}}-\frac{4\left(1+a_{3}\right)}{a_{2} a_{4}} \Longleftrightarrow\left(a_{2}+10\right)\left(10-a_{4}\right)=120$$
It follows from the later equation that $a_{4}<10$. We also have $a_{4} \geq 7$ since the dices are 7 and therefore the minimum sum is 7. It remains to check the possibilities $a_{4}=7,8$ and $9$. The only solution appears for $a_{4}=7$ (the maximum sum is 42) which gives $a_{3}=8$, $a_{2}=30$ and $a_{1}=31$.