Given a positive integer $n \ge 2$, determine the largest positive integer $N$ for which there exist $N+1$ real numbers $a_0, a_1, \dots, a_N$ such that
$(1) \ $   $a_0+a_1 = -\frac{1}{n},$ and
$(2) \ $   $(a_k+a_{k-1})(a_k+a_{k+1})=a_{k-1}-a_{k+1}$ for $1 \le k \le N-1$.
1. **Claim**: The largest positive integer \( N \) for which there exist \( N+1 \) real numbers \( a_0, a_1, \dots, a_N \) satisfying the given conditions is \( \boxed{n} \).

2. **Base Case**: We start by verifying the base cases for \( k = 0 \) and \( k = 1 \).

   - For \( k = 0 \):
     \[
     a_0 + a_1 = -\frac{1}{n}
     \]
     This is given directly by the problem statement.

   - For \( k = 1 \):
     \[
     (a_1 + a_0)(a_1 + a_2) = a_0 - a_2
     \]
     Substituting \( a_0 + a_1 = -\frac{1}{n} \):
     \[
     \left(a_1 - \frac{1}{n} \right)(a_1 + a_2) = -\frac{1}{n} - a_2
     \]
     Let \( a_1 + a_2 = x \). Then:
     \[
     \left(a_1 - \frac{1}{n} \right)x = -\frac{1}{n} - a_2
     \]
     Since \( a_1 = -\frac{1}{n} - a_0 \), we substitute \( a_0 = 0 \) (as a simplification):
     \[
     a_1 = -\frac{1}{n}
     \]
     Thus:
     \[
     \left(-\frac{1}{n} - \frac{1}{n} \right)x = -\frac{1}{n} - a_2
     \]
     \[
     -\frac{2}{n} x = -\frac{1}{n} - a_2
     \]
     Solving for \( x \):
     \[
     x = \frac{1}{1-n}
     \]
     Therefore:
     \[
     a_1 + a_2 = \frac{1}{1-n}
     \]

3. **Inductive Step**: Assume that for some \( k \), \( a_k + a_{k-1} = \frac{1}{k-1-n} \). We need to show that \( a_k + a_{k+1} = \frac{1}{k-n} \).

   - Given:
     \[
     (a_k + a_{k-1})(a_k + a_{k+1}) = a_{k-1} - a_{k+1}
     \]
     Substitute \( a_k + a_{k-1} = \frac{1}{k-1-n} \):
     \[
     \left(\frac{1}{k-1-n}\right)(a_k + a_{k+1}) = a_{k-1} - a_{k+1}
     \]
     Let \( a_k + a_{k+1} = y \):
     \[
     \frac{y}{k-1-n} = \frac{1}{k-1-n} - a_{k+1}
     \]
     Multiply both sides by \( k-1-n \):
     \[
     y = 1 - (k-1-n) a_{k+1}
     \]
     Since \( y = a_k + a_{k+1} \), we have:
     \[
     (k-n)(a_k + a_{k+1}) = 1
     \]
     Therefore:
     \[
     a_k + a_{k+1} = \frac{1}{k-n}
     \]
     This completes the induction.

4. **Conclusion**: If \( N > n \), then \( a_n + a_{n+1} \) becomes undefined, so \( N \le n \). For equality, we can set \( a_0 = 0 \) and recursively define \( a_{i+1} = \frac{1}{i-n} - a_i \) for \( 0 \le i \le n-1 \).