everyone will get it

A normal year has 365 days, but a leap year has 366 days. Therefore there are 366 possible birthdays. Assuming that each 366 people in the group has a different birthday, there is 1 person remaining. Therefore this person will have to share a person with one other person; therefore the minimum number of people with the same birthday is 2.

One year(normal year) has $365$ days .In order to get the least number of people we must assume that each of $365$ people on this group has their birthday on different days.But there are two people left .Certainly,one of these two people must have his birthday on some random day of the year which will necessarily correspond with the birthday of some other person in the $365$ group.Therefore the asnwer is $\boxed{2}$ .