Determine all nonnegative integers $n$ having two distinct positive divisors with the same distance from $\tfrac{n}{3}$.

(Richard Henner)
1. We start by noting that \( n \) must be a positive integer greater than 0, and we need to find \( n \) such that it has two distinct positive divisors with the same distance from \( \frac{n}{3} \).

2. Let \( n = 3k \) for some integer \( k \). This is because if \( n \) is not a multiple of 3, then \( \frac{n}{3} \) is not an integer, and it would be impossible for two divisors to have the same distance from a non-integer.

3. Suppose \( d_1 \) and \( d_2 \) are two distinct positive divisors of \( n \) such that \( d_1 \) and \( d_2 \) are equidistant from \( \frac{n}{3} \). Let \( d_1 = k + a \) and \( d_2 = k - a \) for some positive integer \( a \) where \( 1 \leq a < k \).

4. Since \( d_1 \) and \( d_2 \) are divisors of \( n \), we have:
   \[
   k + a \mid 3k \quad \text{and} \quad k - a \mid 3k
   \]

5. Note that \( k + a \) is a divisor of \( n \) greater than \( k \), so it can only be \( \frac{n}{2} \) or \( n \). Since \( a < k \), we have \( k + a < 2k < n \), so it must be the case that:
   \[
   k + a = \frac{n}{2}
   \]

6. This implies that \( n \) must be even. Let \( n = 6m \) for some integer \( m \). Then:
   \[
   k + a = 3m
   \]

7. Since \( k = 2m \), we have:
   \[
   2m + a = 3m \implies a = m
   \]

8. Therefore, \( k + a = 3m \) is always a divisor of \( n \), and \( k - a = 2m - m = m \) is also always a divisor. Hence, the numbers we are looking for are all positive multiples of 6.

The final answer is \( \boxed{ n } \) is a positive multiple of 6.