Solution 1. Either there are 3 consecutive empty seats or two separate pairs of empty seats. In the first case, there are $2N$ possibilities as the middle of the three empty seats can be any of the $2N$ seats. In the second case, there are two pairs of empty seats. If the seats are numbered 1 to $2N$ then one of the pairs of empty seats will be odd-even and the other even-odd. There are $N$ choices for the odd-even empty pair. For each of these there are $N-2$ choices for the odd number in the even-odd pair. That gives rise to $N(N-2)$ seating arrangements. Adding the two cases gives a total of $N^2$ possibilities.

Solution 2. Divide the $2N$ seats into $N$ pairs of consecutive seats. $N-1$ of these pairs contain one person and the other is empty. There are $N$ ways to choose which pair is empty. The remaining $2N-2$ seats are occupied in the form 01 or 10 where 1 represents occupation. Working clockwise from the empty pair, the sequence starts with a (possibly empty) sequence of the form 101010...10 and finishes with a (possibly empty) sequence of the form 01...0101. Considering the possible lengths of the starting sequence we find that there are $N$ different possible sequences of this kind. This gives a total of $N \times N$ possibilities.

Solution 3. Let $P_n$ be the path graph with $n$ edges. A set of vertices $H$ of a graph $G$ is an independent set if no two vertices of $H$ are adjacent. Let $I(G, m)$ = the number of independent sets of size $m$ of the graph $G$. Number the seats 1, 2, ..., $2N$. There are two options:
(A) seat 1 is occupied and so seats 2 and $2N$ are empty;
(B) seat 1 is empty.
The number of possibilities in case (A) is $I(P_{2N-4}, N-2)$ and in case (B) is $I(P_{2N-2}, N-1)$. Using the ideas from solution 1 or 2 it is easy to show that $I(P_{2k}, k) = \frac{(k+2)(k+1)}{2}$. Hence the number of possibilities is equal to $\frac{N(N-1)}{2} + \frac{(N+1)N}{2} = N^2$.