Let $x_1, x_2, . . . , x_{2022}$ be nonzero real numbers. Suppose that $x_k + \frac{1}{x_{k+1}} < 0$ for each $1 \leq k \leq 2022$, where $x_{2023}=x_1$. Compute the maximum possible number of integers $1 \leq n \leq 2022$ such that $x_n > 0$.
1. **Understanding the Problem:**
   We are given a sequence of nonzero real numbers \( x_1, x_2, \ldots, x_{2022} \) such that \( x_k + \frac{1}{x_{k+1}} < 0 \) for each \( 1 \leq k \leq 2022 \), with \( x_{2023} = x_1 \). We need to find the maximum number of integers \( 1 \leq n \leq 2022 \) such that \( x_n > 0 \).

2. **Analyzing the Condition:**
   The condition \( x_k + \frac{1}{x_{k+1}} < 0 \) implies:
   \[
   x_k < -\frac{1}{x_{k+1}}
   \]
   If \( x_k > 0 \), then \( -\frac{1}{x_{k+1}} \) must be negative, which means \( x_{k+1} \) must be negative. Conversely, if \( x_k < 0 \), then \( -\frac{1}{x_{k+1}} \) must be positive, which means \( x_{k+1} \) must be positive. Therefore, the sequence must alternate in sign.

3. **Counting Positive Numbers:**
   Since the sequence alternates in sign, if we start with a positive number, the sequence will be \( +, -, +, -, \ldots \). If we start with a negative number, the sequence will be \( -, +, -, +, \ldots \).

4. **Maximum Number of Positive Numbers:**
   To maximize the number of positive numbers, we can start with a positive number. Since the sequence alternates, the number of positive numbers will be half of the total number of terms if the total number is even, or one more than half if the total number is odd.

   Given \( 2022 \) terms, the maximum number of positive numbers is:
   \[
   \left\lfloor \frac{2022 + 1}{2} \right\rfloor = \left\lfloor 1011.5 \right\rfloor = 1011
   \]

5. **Verifying the Maximum:**
   We need to verify if having \( 1011 \) positive numbers is possible. If we start with \( x_1 > 0 \), then \( x_2 < 0 \), \( x_3 > 0 \), and so on. The sequence will be:
   \[
   x_1, x_2, x_3, \ldots, x_{2022}
   \]
   with \( x_{2023} = x_1 \). Since \( x_1 \) is positive and the sequence alternates, \( x_{2022} \) will be negative, and \( x_{2023} = x_1 \) will be positive, which is consistent.

6. **Conclusion:**
   Therefore, the maximum possible number of integers \( 1 \leq n \leq 2022 \) such that \( x_n > 0 \) is \( 1011 \).

The final answer is \( \boxed{1011} \).