CoNP and NPhard intersection
There is a problem which is both NP-hard and in coNP if and only if NP = coNP.

If NP = coNP, than NP-complete problems (like 3-SAT) are both NP-hard and in coNP. 

On the other hand, if any NP-hard problem is in coNP, then all problems in NP are reducible to it, so all problems in NP are in coNP so NP &sube; coNP. Now, since the complement of NP is coNP, and vice versa, we also have  coNP &sube; NP. This means NP = coNP. 

The question of whether NP = coNP is open, but most theoretical computer scientists do not think it is very likely.