If $c \geq 1,d \geq 1$ then $a \geq c+1$ so $d+1 \leq b \leq \frac{2(1+cd)}{c+1} \leq 2d$ and same way $c+1 \leq a \leq 2c$ So $a-c \leq c, b-d \leq d$ and $(a-c)+(b-d) \leq c+d \to$ contradiction So let $c=0$ then $ab=2$ and so $(a,b,c,d)=(1,2,0,1)$ For $d=0$ we get $(a,b,c,d)=(2,1,1,0)$

Take WLOG $a \le b$. We have $b \le a > c$. Write $b = ak$ with $k \ge 1$ a rational number. We have $(a-c)+(b-d) \ge c+d \iff a + b \ge 2(c+d) \iff a + ak \ge 2(c+d)$. Notice that $cd \le (a-1)(ak+1)$ clearly, so we have $a^2k = 2(1+cd) \ge 2(1+(a-1)(ak+1)) = 2(a^2k+a-ak)$, so we get that $ak < 2k-2$ which means for $a \ge 2$ no solution. Therefore check $a=1$, so $c=0$ and so $b=2$ and $d=1$, therefore $(a,b,c,d) = (1,2,0,1)$ or $(a,b,c,d) = (2,1,1,0)$ $\square$.