Solution Suppose each person books 2 seats: the chair they want to sit on and the chair to its left. Then there are only two chairs remaining once everyone has booked their seats. Consider the possible cases for how many chairs are between the two remaining chairs. If there is an odd number of chairs between the two remaining chairs, then we have a contradiction unless the remaining chairs are beside each other (as otherwise there would need to be an extra empty chair). If there is an even number of chairs between the remaining chairs, then we can book everyone in the pairs of non-remaining chairs. If we label the seats from $1$ to $2N$, then we can choose either 2 even numbered seats, two odd numbered seats or two seats right next to each other to be left unbooked. There is $\binom{N}{2}$ ways to pick to even/odd numbered seats, and $N$ ways to choose two seats next to each other, so there is $$ \binom{N}{2} + \binom{N}{2} + N = N(N - 1) + N = N^2, $$ways to seat the guests.