Determine all positive integers $n{}$ which can be expressed as $d_1+d_2+d_3$ where $d_1,d_2,d_3$ are distinct positive divisors of $n{}$.
1. Let \( d_1, d_2, d_3 \) be distinct positive divisors of \( n \) such that \( d_1 < d_2 < d_3 \). Then, we have:
   \[
   d_1 + d_2 + d_3 = n
   \]
   Since \( d_1, d_2, d_3 \) are distinct divisors of \( n \), it follows that \( d_3 < n \).

2. Given \( d_1 < d_2 < d_3 \), we can infer:
   \[
   d_1 + d_2 + d_3 < 3d_3
   \]
   Therefore:
   \[
   n < 3d_3
   \]
   Since \( d_3 < n \), we can write:
   \[
   d_3 < n < 3d_3
   \]
   This implies:
   \[
   n = 2d_3
   \]
   Hence, we have:
   \[
   d_1 + d_2 + d_3 = 2d_3
   \]
   Simplifying, we get:
   \[
   d_1 + d_2 = d_3
   \]

3. Next, we consider the inequality:
   \[
   4d_2 > 2(d_1 + d_2) = d_1 + d_2 + d_3 = n > 2d_2
   \]
   This implies:
   \[
   n = 3d_2
   \]
   Therefore:
   \[
   3d_2 = 2d_3 = 2(d_1 + d_2)
   \]
   Simplifying, we get:
   \[
   3d_2 = 2d_1 + 2d_2
   \]
   Subtracting \( 2d_2 \) from both sides, we obtain:
   \[
   d_2 = 2d_1
   \]
   Substituting \( d_2 = 2d_1 \) into \( d_3 = d_1 + d_2 \), we get:
   \[
   d_3 = d_1 + 2d_1 = 3d_1
   \]
   Therefore:
   \[
   n = 6d_1
   \]

4. For every \( n = 6k \), we can choose \( d_1 = k \), \( d_2 = 2k \), and \( d_3 = 3k \). These values satisfy the conditions \( d_1 + d_2 + d_3 = n \) and \( d_1, d_2, d_3 \) are distinct positive divisors of \( n \).

The final answer is \( \boxed{ n = 6k } \) for any positive integer \( k \).