Hash function that allows to decide if A > B if you only have hash(A) and hash(B)? Is it possible to construct a hash function with the following property?

If you have hash(A) and hash(B) with A and B being integers, you can tell if A is greater than B -- without however knowing the actual values of A and B.

Even better: If you have hash(A) and hash(B) with A and B being ordered sets of strings, you can tell which elements of A overlap with B (e.g. those at position 1 and 3 in the ordered set) -- without however knowing the actual contents of A and B.
For hash function $h : \{0,1\}^* \rightarrow \{0,1\}^k$, this is not possible. This is because there are more possible inputs than outputs (pigeon hole principle). And this means that for some $A &lt; B$ we have $h(A) = h(B)$. Thus there will be no way to tell the order of $A$ and $B$.

Addition:  To address some of the comments; note that this answer only talks about hash functions defined as functions where the domain is the set of all bit strings and the codomain a set of bit strings of some fixed length $k$. It also assumes that inputs $A$ and $B$ are interpreted as integers (as specified in the question). 

The argument only uses that such a hash function must have collisions. We can in fact generalize the argument to any hash function with collisions (i.e., also functions with different domain and codomains). Note that the argument only assumes the existence of collisions and thus holds even if the hash function is collision resistant.   

The argument says that for such hash functions there will be some pairs of inputs, those defining a hash function collision, where the order cannot be decided from the hash alone. What happens on other inputs pairs the argument does not directly say anything about. 

It is true however, as some comments point out, that if we define hash functions to include functions completely without collisions, then the argument does not hold for all hash functions. However, for functions where my argument does not apply we can use the argument of fgrieu, to show that a hash function with the desired properties can at least not be pre-image resistant.