1. (a) For which positive integers n, does 2n divide the sum of the first n positive
integers?
(b) Determine, with proof, those positive integers n (if any) which have the
property that 2n + 1 divides the sum of the first n positive integers.
### Part (a)
1. The sum of the first \( n \) positive integers is given by the formula:
   \[
   S = \frac{n(n+1)}{2}
   \]
2. We need to determine for which positive integers \( n \), \( 2n \) divides \( S \). This means:
   \[
   2n \mid \frac{n(n+1)}{2}
   \]
3. Simplify the expression:
   \[
   2n \mid \frac{n(n+1)}{2} \implies 4n \mid n(n+1)
   \]
4. Since \( n \) is a factor of \( n(n+1) \), we need:
   \[
   4 \mid (n+1)
   \]
5. Therefore, \( n+1 \) must be a multiple of 4:
   \[
   n+1 = 4k \quad \text{for some integer } k
   \]
6. Solving for \( n \):
   \[
   n = 4k - 1
   \]
7. Thus, \( n \) must be of the form \( 4k - 1 \) for some integer \( k \).

### Part (b)
1. The sum of the first \( n \) positive integers is again:
   \[
   S = \frac{n(n+1)}{2}
   \]
2. We need to determine for which positive integers \( n \), \( 2n + 1 \) divides \( S \). This means:
   \[
   2n + 1 \mid \frac{n(n+1)}{2}
   \]
3. Simplify the expression:
   \[
   2(2n + 1) \mid n(n+1)
   \]
4. Since \( 2n + 1 \) is a factor of \( n(n+1) \), we need:
   \[
   2n + 1 \mid n(n+1)
   \]
5. Consider the expression \( n(n+1) \mod (2n + 1) \):
   \[
   n(n+1) \equiv 0 \pmod{2n + 1}
   \]
6. Let \( k = 2n + 1 \), then \( n = \frac{k-1}{2} \):
   \[
   \left(\frac{k-1}{2}\right)\left(\frac{k-1}{2} + 1\right) \equiv 0 \pmod{k}
   \]
7. Simplify the expression:
   \[
   \left(\frac{k-1}{2}\right)\left(\frac{k+1}{2}\right) \equiv 0 \pmod{k}
   \]
8. This simplifies to:
   \[
   \frac{(k-1)(k+1)}{4} \equiv 0 \pmod{k}
   \]
9. Since \( (k-1)(k+1) = k^2 - 1 \):
   \[
   \frac{k^2 - 1}{4} \equiv 0 \pmod{k}
   \]
10. This implies:
    \[
    k^2 - 1 \equiv 0 \pmod{4k}
    \]
11. Since \( k^2 - 1 \equiv 0 \pmod{k} \), we need:
    \[
    k^2 - 1 \equiv 0 \pmod{4}
    \]
12. This implies \( k \equiv 1 \pmod{4} \) or \( k \equiv 3 \pmod{4} \).

Therefore, the positive integers \( n \) which have the property that \( 2n + 1 \) divides the sum of the first \( n \) positive integers are those for which \( 2n + 1 \equiv 1 \pmod{4} \) or \( 2n + 1 \equiv 3 \pmod{4} \).

The final answer is \( \boxed{ n = 4k - 1 } \) for part (a) and \( 2n + 1 \equiv 1 \pmod{4} \) or \( 2n + 1 \equiv 3 \pmod{4} \) for part (b).