Since everything is cyclic, w.l.o.g. $a=\max(a,b,c,d)$. So $\frac{b+c}{a} \in \{1,2\}$. If $b+c=2a$, then $a=b=c$ and hence $a \mid d$ so that $\boxed{a=b=c=d=1}$ which is indeed a solution. Now suppose that $b+c=a$. First case: $b=\max(b,c,d)$. Then $\frac{c+d}{b} \in \{1,2\}$. If $c+d=2b$, then $\boxed{a=2, b=c=d=1}$ which is indeed a solution. If $c+d=b$, then $c \mid 2d, d \mid 3c$ and $(c;d)=1$. Hence $c \mid 2, d \mid 3$. So we get the solutions $\boxed{(3,2,1,1), (5,3,2,1), (5,4,1,3),(7,5,2,3)}$ all of them are indeed solutions. Second case: $c=\max(b,c,d)$ and $c>b$. Then $b+d=c$ and hence $b \mid 2d$ and $d \mid 3b$ an $(b;d)=1$, hence $b \mid 2, d \mid 3$. So we get the solutions $\boxed{(3,1,2,1), (5,1,4,3), (5,2,3,1), (7,2,5,3)}$. Third case: $d>b,c$. Then $\frac{2b+c}{d} \in \{1,2\}$. But $d<a=b+c$, hence $2b+c=2d$. Hence $c$ is even and $c \mid 4b$ and $b \mid 3c$ and so $c \mid 4, b \mid 3$ and we get the additional solution $\boxed{(7,3,4,5)}$. And of course all the permutations...