Let $1 = d_1 < d_2 < ...< d_k = n$ be all natural divisors of the natural number $n$. Find all possible values ​​of the number $k$ if  $n=d_2d_3 + d_2d_5+d_3d_5$.
We are given a natural number \( n \) with divisors \( 1 = d_1 < d_2 < \cdots < d_k = n \) satisfying the equation \( n = d_2d_3 + d_2d_5 + d_3d_5 \).

We need to find all possible values of \( k \).

### Case 1: \( (d_2, d_3) = (p, p^2) \)

1. Let \( d_2 = p \), where \( p \) is the smallest prime divisor of \( n \).
2. Then \( d_3 = p^2 \).
3. From the given equation \( n = d_2d_3 + d_2d_5 + d_3d_5 \), we have:
   \[
   n = p \cdot p^2 + p \cdot d_5 + p^2 \cdot d_5
   \]
   \[
   n = p^3 + p \cdot d_5 + p^2 \cdot d_5
   \]
   \[
   n = p^3 + d_5(p + p^2)
   \]
4. Since \( d_5 \mid p^3 \) and \( d_5 > p^2 \), we get \( d_5 = p^3 \).
5. Substituting \( d_5 = p^3 \) into the equation, we get:
   \[
   n = p^3 + p \cdot p^3 + p^2 \cdot p^3
   \]
   \[
   n = p^3 + p^4 + p^5
   \]
   \[
   n = p^3(p^2 + p + 1)
   \]
6. We know \( d_5 = p^3 \) and \( d_5 > \sqrt{n} \), so:
   \[
   p^3 > \sqrt{p^3(p^2 + p + 1)}
   \]
   \[
   p^3 > p^{3/2} \sqrt{p^2 + p + 1}
   \]
   \[
   p^{3/2} > \sqrt{p^2 + p + 1}
   \]
   \[
   p^3 > p^2 + p + 1
   \]
7. This implies \( k < 10 \).

### Case 2: \( (d_2, d_3) = (p, q) \)

1. Let \( d_2 = p \) and \( d_3 = q \), where \( p \) and \( q \) are the smallest and second smallest prime divisors of \( n \), respectively.
2. From the given equation \( n = d_2d_3 + d_2d_5 + d_3d_5 \), we have:
   \[
   n = p \cdot q + p \cdot d_5 + q \cdot d_5
   \]
   \[
   n = pq + d_5(p + q)
   \]
3. Since \( d_5 \mid pq \) and \( d_5 > q \), we get \( d_5 = pq \).
4. Substituting \( d_5 = pq \) into the equation, we get:
   \[
   n = pq + pq(p + q)
   \]
   \[
   n = pq(1 + p + q)
   \]
5. We know \( d_3 < d_4 < d_5 \), so:
   \[
   q < d_4 < pq
   \]
6. This implies \( d_4 = p + q + 1 \) when \( (p, q) \neq (2, 3) \).
7. Consequently, \( n = d_5d_4 \), which implies \( k = 4 + 5 - 1 = 8 \).

### Special Case: \( (p, q) = (2, 3) \)

1. Setting \( (p, q) = (2, 3) \) in the equation, we get:
   \[
   n = 2 \cdot 3 + 2 \cdot 6 + 3 \cdot 6
   \]
   \[
   n = 6 + 12 + 18
   \]
   \[
   n = 36
   \]
2. The divisors of 36 are \( 1, 2, 3, 4, 6, 9, 12, 18, 36 \), so \( k = 9 \).

### Conclusion

The natural numbers \( n \) satisfying the given condition have 8 or 9 divisors, i.e., \( k \in \{8, 9\} \).

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