jmc

We have $$\begin{array}{rl}abcd& =1000a+100b+10c+d,\text{ and }\\ dcba& =1000d+100c+10b+a\end{array}$$ Adding these gives $$\begin{array}{rl}abcd+dcba& =(1000+1)d+(100+10)c\\ & +(10+100)b+(1+1000)a\\ & =1001(a+d)+110(b+c).\end{array}$$ Furthermore, since $a,b,c,d$ are consecutive, we have $b=a+1$, $c=a+2$, and $d=a+3$, so that $$a+d=2a+3=b+c.$$ Hence, $$abcd+dcba=1001(2a+3)+110(2a+3)=1111(2a+3).$$ It follows that $\boxed{1111}$ must divide any number of the given form. To see that no higher number must divide it, if we take $a=1$ and $a=2$, we get the numbers $5555$ and $7777$, whose greatest common factor is indeed $1111$.