A round table has $2 N$ chairs around it. Due to social distancing guidelines, no two people are allowed to sit next to each other. How many different ways are there to choose seats around the table on which $N-1$ guests can be seated?
1. **Understanding the Problem:**
   We have a round table with \(2N\) chairs, and we need to seat \(N-1\) guests such that no two guests are seated next to each other. This means there must be at least one empty chair between any two guests.

2. **Initial Consideration:**
   To ensure no two guests sit next to each other, we can think of each guest as occupying two chairs: the one they sit on and the one immediately to their left (or right, but we will stick to one direction for consistency). This effectively means each guest "blocks" two chairs.

3. **Counting the Remaining Chairs:**
   If \(N-1\) guests are seated, they will occupy \(2(N-1)\) chairs. This leaves \(2N - 2(N-1) = 2N - 2N + 2 = 2\) chairs unoccupied.

4. **Arranging the Remaining Chairs:**
   We need to consider the arrangement of these remaining two chairs. Since the table is round, the position of these two chairs relative to each other is crucial. We need to ensure that the remaining chairs are not next to each other, as this would violate the seating rule.

5. **Case Analysis:**
   - **Case 1: Even Number of Chairs Between Remaining Chairs:**
     If there is an even number of chairs between the two remaining chairs, then the guests can be seated in pairs of non-remaining chairs. This ensures that no two guests are next to each other.
   - **Case 2: Odd Number of Chairs Between Remaining Chairs:**
     If there is an odd number of chairs between the two remaining chairs, this would imply that there is an extra empty chair somewhere, which contradicts the seating rule unless the remaining chairs are next to each other.

6. **Labeling the Seats:**
   Label the seats from \(1\) to \(2N\). We need to choose two seats to remain unoccupied. These two seats can either both be even-numbered, both be odd-numbered, or be next to each other.

7. **Counting the Ways:**
   - **Choosing Two Even-Numbered Seats:**
     There are \(\binom{N}{2}\) ways to choose two even-numbered seats.
   - **Choosing Two Odd-Numbered Seats:**
     Similarly, there are \(\binom{N}{2}\) ways to choose two odd-numbered seats.
   - **Choosing Two Adjacent Seats:**
     There are \(N\) ways to choose two adjacent seats.

8. **Total Number of Ways:**
   Summing up all the possible ways, we get:
   \[
   \binom{N}{2} + \binom{N}{2} + N = \frac{N(N-1)}{2} + \frac{N(N-1)}{2} + N = N(N-1) + N = N^2
   \]

Conclusion:
\[
\boxed{N^2}
\]