A sequence of $N$ consecutive positive integers is called [i]good [/i] if it is possible to choose two of these numbers so that their product is divisible by the sum of the other $N-2$ numbers. For which $N$ do there exist infinitely many [i]good [/i] sequences?
1. **Define the sequence and the problem condition:**
   A sequence of \( N \) consecutive positive integers is called *good* if it is possible to choose two of these numbers so that their product is divisible by the sum of the other \( N-2 \) numbers. We need to determine for which \( N \) there exist infinitely many *good* sequences.

2. **Consider the case for even \( N \geq 4 \):**
   Let the sequence be \( a+1, a+2, a+3, \ldots, a+N \).

3. **Sum of \( N-2 \) numbers:**
   The sum of the \( N-2 \) numbers, excluding \( a+\frac{N}{2} \) and \( a+N \), is:
   \[
   S = \sum_{i=1}^{N} (a+i) - (a+\frac{N}{2}) - (a+N) = aN + \frac{N(N+1)}{2} - (a+\frac{N}{2}) - (a+N)
   \]
   Simplifying the sum:
   \[
   S = aN + \frac{N(N+1)}{2} - a - \frac{N}{2} - a - N = a(N-2) + \frac{N(N+1)}{2} - N - \frac{N}{2}
   \]

4. **Divisibility condition:**
   We need:
   \[
   a(N-2) + \frac{N(N+1)}{2} - N - \frac{N}{2} \mid (a+\frac{N}{2})(a+N)
   \]
   Simplifying the divisibility condition:
   \[
   N-2 \mid a+N
   \]
   This implies:
   \[
   a = (N-2)k - N
   \]
   where \( k \) can take infinitely many values.

5. **Conclusion for even \( N \geq 4 \):**
   Therefore, for even \( N \geq 4 \), there exist infinitely many *good* sequences.

6. **Consider the case for odd \( N \geq 3 \):**
   Let the sequence be \( a-\frac{N-1}{2}, a-\frac{N-3}{2}, \ldots, a+0, \ldots, a+\frac{N-3}{2}, a+\frac{N-1}{2} \).

7. **Assume the divisibility condition holds:**
   For the sake of contradiction, assume:
   \[
   a(N-2) - i - j \mid (a+i)(a+j)
   \]
   This simplifies to:
   \[
   a(N-2) - i - j \mid ij(N-2)^2 + (i+j)^2(N-1)
   \]

8. **Analyze the simplified condition:**
   For this to hold infinitely many times, we must have:
   \[
   (i+j)^2(N-1) = -ij(N-2)^2
   \]
   Simplifying further:
   \[
   i^2(N-1) + i(j(N^2-2N+2)) + j^2(N-1) = 0
   \]
   Solving for \( i \):
   \[
   i = \frac{-j(N^2-2N+2) \pm \sqrt{j^2(N^2-2N+2)^2 - 4(N-1)^2j^2}}{2(N-1)}
   \]
   This implies:
   \[
   i = -j \cdot (N-1) \quad \text{or} \quad i = -j \cdot \frac{1}{N-1}
   \]
   Therefore, \( N-1 \mid i \) or \( j \), leading to \( i = j = 0 \), which violates the conditions.

9. **Conclusion for odd \( N \geq 3 \):**
   Hence, for odd \( N \geq 3 \), there do not exist infinitely many *good* sequences.

The final answer is \(\boxed{\text{All even numbers } N \geq 4}\).