Six real numbers $x_1<x_2<x_3<x_4<x_5<x_6$ are given. For each triplet of distinct numbers of those six Vitya calculated their sum. It turned out that the $20$ sums are pairwise distinct; denote those sums by $$s_1<s_2<s_3<\cdots<s_{19}<s_{20}.$$ It is known that $x_2+x_3+x_4=s_{11}$, $x_2+x_3+x_6=s_{15}$ and $x_1+x_2+x_6=s_{m}$. Find all possible values of $m$.
1. **Identify the sums involving \( x_1 \) and \( x_6 \):**
   - Given \( x_2 + x_3 + x_4 = s_{11} \) and \( x_2 + x_3 + x_6 = s_{15} \), we know that \( s_{11} \) is the smallest sum that does not include \( x_1 \).
   - Since \( s_{15} \) is the smallest sum involving \( x_6 \) among the sums that do not include \( x_1 \), any sum containing \( x_6 \) must be greater than any sum not containing \( x_6 \).

2. **Compare sums involving \( x_6 \) and those not involving \( x_6 \):**
   - From \( s_{15} = x_2 + x_3 + x_6 \) and \( s_{11} = x_2 + x_3 + x_4 \), we have:
     \[
     x_2 + x_3 + x_6 > x_2 + x_3 + x_4 \implies x_6 > x_4
     \]
   - Since \( x_2 + x_3 + x_6 \) is the smallest sum involving \( x_6 \), any sum involving \( x_6 \) must be greater than any sum not involving \( x_6 \).

3. **Determine the position of \( x_1 + x_2 + x_6 \):**
   - We know \( x_2 + x_3 + x_4 = s_{11} \) and \( x_2 + x_3 + x_6 = s_{15} \).
   - Since \( x_1 < x_2 \), any sum involving \( x_1 \) should be smaller than \( x_2 + x_3 + x_4 \) (because there are exactly 10 sums that contain \( x_1 \), and \( s_{11} \) is the smallest sum that does not contain \( x_1 \)).

4. **Analyze the sums involving \( x_1 \):**
   - There are 10 sums involving \( x_1 \). Since \( x_1 < x_2 \), these sums are smaller than \( x_2 + x_3 + x_4 \).
   - The sums involving \( x_1 \) and \( x_6 \) are:
     \[
     x_1 + x_2 + x_6, \quad x_1 + x_3 + x_6, \quad x_1 + x_4 + x_6, \quad x_1 + x_5 + x_6
     \]
   - Since \( x_1 + x_2 + x_6 \) is smaller than \( x_1 + x_3 + x_6 \), \( x_1 + x_4 + x_6 \), and \( x_1 + x_5 + x_6 \), it must be among the smallest sums involving \( x_1 \).

5. **Determine the exact position of \( x_1 + x_2 + x_6 \):**
   - Given \( x_1 + x_2 + x_6 \) is smaller than \( x_1 + x_3 + x_6 \), \( x_1 + x_4 + x_6 \), and \( x_1 + x_5 + x_6 \), it must be the 7th smallest sum:
     \[
     s_7 = x_1 + x_2 + x_6
     \]

Thus, the possible value of \( m \) is \( 7 \).

The final answer is \( \boxed{7} \)