Without even trying to hard the answer is use the pigeon hole principle! -- edit sorry this should say injectivity.
The answer is n-1. The logic used to it is rather tedious. Here is a quick overview.
There are 2n people. First, we must use injectivity. 2n-1 people reported 2n-1 different number of handshakes. Since the maximum number of handshakes is 2n-1-1 = 2n-2 a person can make, that means each persons number of handshakes is between 0 and 2n-2. Since there are 2n-1 values between 0 and 2n-2 we must have by injectivity that each person other than the hostess must have a unique number of handshakes between 0 and 2n-2. Label each person (other than the hostess) by their number of handshakes.
Consider person 2n-2, this person must have handshook everyone except 0, including the hostess. This is because there are 2n-1 possible people other than 2n-2 himself, but 0 is not available. Then consider person 2n-3 this person can not handshake 0 or 1 (1 has reached his maximum allowed handshakes) thus he must have handshook everyone except those 2. Continue this process until you reach person n this person must handshake everyone except people 0, ..., n-2. Now consider person n-1. This person has reached their exact allowed quota of handshakes already. This is true of all people n-1 to 0. Now consider the hostess. She can not be allowed to handshake any additional people for every person has reached their quota of handshakes. Thus she has handshook everyone from 2n-2 to n that is 2n-2 -n + 1 = n -1 people.