Find all integers $k \ge 5$ for which there is a positive integer $n$ with exactly $k$ positive divisors
$1 = d_1 <d_2 < ... <d_k = n$ and $d_2d_3 + d_3d_5 + d_5d_2 = n$.
1. Let \( p \) be the smallest prime divisor of \( n \). From the given equation, since \( d_2 \mid n \), \( d_3 \mid n \), and \( d_5 \mid n \), it follows that \( d_2 \mid d_3d_5 \), \( d_3 \mid d_2d_5 \), and \( d_5 \mid d_2d_3 \). In particular, \( d_2 = p \mid d_3d_5 \).

2. **Case 1: \( p \mid d_3 \)**. Notice that \( d_3 = pq \) for \( q \neq p \) and \( q \) prime immediately yields a contradiction. So, \( d_3 = p^2 \). Moreover, we have that \( d_5 \mid d_2d_3 = p^3 \), so \( d_5 = p^3 \). Thus, \( n = d_2d_3 + d_3d_5 + d_5d_2 = p^3(p^2 + p + 1) \).

3. Now since for any prime \( p \) the inequality \( p^2 < p^2 + p + 1 < p^3 \) clearly holds and \( p^2 + p + 1 \mid n \), \( d_4 \) must be \( p^2 + p + 1 \).

4. We have found two divisors \( d_4 \) and \( d_5 \) such that \( d_4d_5 = n \). From here, the list of divisors of \( n \) (in ascending order) is:
   \[
   1, p, p^2, p^2 + p + 1, p^3, p(p^2 + p + 1), p^2(p^2 + p + 1), p^3(p^2 + p + 1)
   \]
   It can be readily seen that \( p^2 + p + 1 \) must be prime and \( n = p^3(p^2 + p + 1) \) with \( p \) and \( p^2 + p + 1 \) prime satisfies the conditions.

5. **Case 2: \( p \mid d_5 \)**. Observe that we get a contradiction if \( d_5 \) and \( d_2 \) differ by at least three prime factors. If they differ by two prime factors, we again get a contradiction (just check manually the possible subcases). Since \( p \mid d_5 \), \( d_5 = p^2 \) or \( d_5 = pq \) for some prime \( q \neq p \). If \( d_5 = p^2 \), \( d_3 \mid d_5d_2 = p^3 \), so \( d_3 = p \) or \( d_3 = p^2 \), contradiction. Therefore, \( d_5 = pq \) for a prime \( q \neq p \). Now, \( d_3 \mid d_5d_2 = p^2q \) and \( d_5 = pq \mid pd_3 = d_2d_3 \), \( q \mid d_3 \) and \( d_3 = q \). Then, \( n = d_2d_3 + d_2d_5 + d_3d_5 = pq(p + q + 1) \), so \( d_4 = p + q + 1 \) unless \( (p, q) \) is \( (2, 3) \). Note, however, that for \( (p, q) = (2, 3) \), \( n = 36 \) satisfies the condition. For all other prime pairs \( (p, q) \), \( p + q + 1 < pq \). Therefore, as we did in Case 1, there are exactly eight divisors of \( n \) so \( p + q + 1 \) must be prime. Hence, similar to the above case, \( n = pq(p + q + 1) \) with \( p \), \( q \), and \( p + q + 1 \) prime satisfies the conditions.

6. In conclusion, all numbers satisfying our condition are \( n = 36 \), \( n = p^3(p^2 + p + 1) \) ( \( p \) and \( p^2 + p + 1 \) are prime) and \( n = pq(p + q + 1) \) ( \( p \), \( q \), and \( p + q + 1 \) are prime). Therefore, all possible values of \( k \) are \( 8 \), \( 9 \).

The final answer is \( \boxed{ 8, 9 } \)