Certainly! Here is a test case generator for the given problem statement of finding the N-th Lucas number:

```python
import random

def tcgen():
    N = random.randint(1, 86)
    return N

# Function to compute the N-th Lucas number
def compute_lucas_number(N):
    if N == 0:
        return 2
    elif N == 1:
        return 1
    else:
        L0, L1 = 2, 1
        for i in range(2, N + 1):
            L0, L1 = L1, L0 + L1
        return L1

# Generate a test case
N = tcgen()
print(f"Input: {N}")
print(f"Output: {compute_lucas_number(N)}")
```

### Explanation:
1. **Test Case Generator (`tcgen`)**: 
   - It generates a random integer \( N \) within the constraints \( 1 \leq N \leq 86 \).

2. **Lucas Number Calculation (`compute_lucas_number`)**: 
   - It calculates the N-th Lucas number using an iterative approach to avoid recursion limits and stack overflow issues for larger \( N \).

3. **Example of Generated Test Case**:
   - The `tcgen` function generates the input \( N \).
   - The `compute_lucas_number` function computes the N-th Lucas number.
   - The input and output are printed as required for the test case.

### Sample Output:
The script will print a random test case input and its corresponding output, for example:
```
Input: 15
Output: 843
```

This shows the generated input \( N = 15 \) and the computed 15th Lucas number, which is 843.