Here is a test case generator for the given problem statement, which generates the N-th Lucas number:

```python
import random

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

# Example Usage
if __name__ == "__main__":
    N = tcgen()
    print(N)
```

### Explanation:

1. **N Generation**: The generator function `tcgen()` generates a random integer \( N \) within the given constraint \( 1 \leq N \leq 86 \). This range is ensured using `random.randint(1, 86)`.
2. **Output**: The function returns the randomly generated \( N \).

### Example Usage:

When you call `tcgen()`, it will produce a random \( N \). For example:

```python
N = tcgen()
print(N)
```

This will print a random integer \( N \) within the range \([1, 86]\), suitable for testing the problem of finding the N-th Lucas number.

To actually compute the N-th Lucas number for verification or further testing, you can use the following function:

```python
def nth_lucas_number(N):
    if N == 0:
        return 2
    elif N == 1:
        return 1
    else:
        L0, L1 = 2, 1
        for _ in range(2, N + 1):
            L_next = L0 + L1
            L0, L1 = L1, L_next
        return L1

# Example Usage
if __name__ == "__main__":
    N = tcgen()
    print("Generated N:", N)
    print("N-th Lucas Number:", nth_lucas_number(N))
```

### Explanation of the Lucas Number Calculation:

1. **Base Cases**: For \( N = 0 \) and \( N = 1 \), the Lucas numbers are 2 and 1, respectively.
2. **Iterative Calculation**: For \( N \geq 2 \), the Lucas number is calculated iteratively using the relation \( L_i = L_{i-1} + L_{i-2} \).

This will generate the required Lucas number for the given \( N \).