4, 1, 6 and 2

First, set $a-b=n, \frac{a}{b}=m$ . Then, $c-d=m, \frac{c}{d}=n$ .

Since $a=b+n, m=\frac{a}{b}=\frac{b+n}{b}.$

$mb=b+n, b=\frac{n}{m-1}, a=\frac{n}{m-1}+n.$

Do the same way with $c$ and $d$ , then $d=\frac{m}{n-1}, c=\frac{m}{n-1}+m.$

Since $a \neq b \neq c \neq d,$ $m \neq n.$

$a, b, c, d$ are all positive integers. So $n \equiv 0 (\mod{m-1}), m \equiv 0 (\mod{n-1}).$

And $n \geq m-1, m \geq n-1.$

Then, $m-1 \leq n \leq m+1,$ but since $m \neq n,$ $n=m-1 \quad \text{or} \quad n=m+1.$

Let $n$ be $m-1$ . Then, $n+1 \equiv 0 (\mod{n-1})$ .

When $p$ is a positive integer, $n+1=p(n-1).$ $pn-p=n+1, (n-1)(p-1)=2.$

So, $n=2, p=3 \quad \text{or} \quad n=3, p=2.$ However, if $n=2, a=d=3.$

Thus, $n$ should be $3$ , and therefore the only solution for the equation above is $a=4, b=1, c=6, d=2.$