Find the number of positive integers $n$ for which 

(i) $n \leq 1991$; 

(ii) 6 is a factor of $(n^2 + 3n +2)$.
1. We start with the given conditions:
   - \( n \leq 1991 \)
   - \( 6 \) is a factor of \( n^2 + 3n + 2 \)

2. We need to find the number of positive integers \( n \) such that \( n^2 + 3n + 2 \) is divisible by \( 6 \). 

3. First, we factorize the quadratic expression:
   \[
   n^2 + 3n + 2 = (n+1)(n+2)
   \]

4. For \( (n+1)(n+2) \) to be divisible by \( 6 \), it must be divisible by both \( 2 \) and \( 3 \).

5. Consider divisibility by \( 2 \):
   - Since \( n \) is an integer, either \( n \) is even or odd.
   - If \( n \) is even, \( n+1 \) is odd and \( n+2 \) is even.
   - If \( n \) is odd, \( n+1 \) is even and \( n+2 \) is odd.
   - Therefore, \( (n+1)(n+2) \) is always even, satisfying the divisibility by \( 2 \).

6. Consider divisibility by \( 3 \):
   - We need \( (n+1)(n+2) \equiv 0 \pmod{3} \).
   - This implies that at least one of \( n+1 \) or \( n+2 \) must be divisible by \( 3 \).
   - Therefore, \( n \) must not be congruent to \( 0 \pmod{3} \).

7. We now count the number of integers \( n \) from \( 1 \) to \( 1991 \) that are not multiples of \( 3 \):
   - The total number of integers from \( 1 \) to \( 1991 \) is \( 1991 \).
   - The number of multiples of \( 3 \) in this range is \( \left\lfloor \frac{1991}{3} \right\rfloor = 663 \).

8. Subtracting the number of multiples of \( 3 \) from the total number of integers gives:
   \[
   1991 - 663 = 1328
   \]

Conclusion:
The number of positive integers \( n \) for which \( n \leq 1991 \) and \( 6 \) is a factor of \( n^2 + 3n + 2 \) is \( \boxed{1328} \).