Determine all real numbers A such that every sequence of non-zero real numbers $x_1, x_2, \ldots$ satisfying
\[ x_{n+1}=A-\frac{1}{x_n} \]
for every integer $n \ge 1$, has only finitely many negative terms.
To determine all real numbers \( A \) such that every sequence of non-zero real numbers \( x_1, x_2, \ldots \) satisfying
\[ x_{n+1} = A - \frac{1}{x_n} \]
for every integer \( n \ge 1 \), has only finitely many negative terms, we will analyze the behavior of the sequence under different conditions for \( A \).

1. **Case 1: \( A \geq 2 \)**

   - **Observation**: If \( x_i \) is negative, then \( x_{i+1} = A - \frac{1}{x_i} \) will be positive because \( A \geq 2 \) and \( -\frac{1}{x_i} \) is positive and less than 1 in magnitude. Therefore, \( x_{i+1} \) will be greater than \( A - 1 \geq 1 \), ensuring it is positive.

   - **Subcase 1.1: \( x_1 \geq A \)**

     If \( x_1 \geq A \), then:
     \[
     x_2 = A - \frac{1}{x_1} \leq A - \frac{1}{A} \leq A - \frac{1}{2} \leq 2 - \frac{1}{2} = 1.5
     \]
     Since \( x_1 \geq A \geq 2 \), \( x_2 \) will be positive and less than or equal to \( 1.5 \).

     - **Inductive Argument**: If \( x_n \geq 1 \), then \( x_{n+1} = A - \frac{1}{x_n} \geq A - 1 \geq 1 \). Thus, \( x_n \) will never be less than 1, ensuring all terms are positive.

   - **Subcase 1.2: \( x_1 < 1 \)**

     If \( x_1 < 1 \), then:
     \[
     x_2 = A - \frac{1}{x_1} > A - 1 \geq 1
     \]
     This means \( x_2 \) will be positive and greater than or equal to 1, leading us back to the situation in Subcase 1.1.

2. **Case 2: \( 0 \leq A < 2 \)**

   - Choose \( x_1 = 1 \). Then:
     \[
     x_2 = A - 1
     \]
     If \( A < 1 \), \( x_2 \) will be negative. If \( A \geq 1 \), \( x_2 \) will be non-negative but less than 1.

   - If \( x_2 \) is negative, then \( x_3 = A - \frac{1}{x_2} \) will be positive and greater than \( A \). However, since \( A < 2 \), \( x_3 \) will be less than 2.

   - The sequence will oscillate between positive and negative values, ensuring infinitely many negative terms.

3. **Case 3: \( A < 0 \)**

   - If \( A < 0 \), choose \( x_1 = 1 \). Then:
     \[
     x_2 = A - 1 < 0
     \]
     This means \( x_2 \) is negative. The sequence will continue to oscillate between positive and negative values, ensuring infinitely many negative terms.

From the above analysis, we conclude that the only values of \( A \) for which the sequence has only finitely many negative terms are those where \( A \geq 2 \).

The final answer is \( \boxed{ A \geq 2 } \)