We first show that among any 5 integers, sum of 3 of them is divisible by 3. The residue classes modulo 3 are $, , $. By Pigeonhole principle, we have two cases:
- each of these residue classes must have atleast one of the five integers, or
- one residue class must have 3 of these 5 integers belonging to it.
In the former case, let $x_0 \equiv 0 \mod 3$, $x_1 \equiv 1 \mod 3$ and $x_2 \equiv 2 \mod3$. Then summing $x_1, x_2, x_3$, we get, $x_1 + x_2 + x_3 \equiv 0 \mod 3$.
In the latter case, we have 3 integers among the 5, say $x_1, x_2, x_3$ such that, $x_1 \equiv x_2 \equiv x_3 \equiv k \mod 3$, again summing these three we get $x_1 + x_2 + x_3 \equiv 3k \equiv 0 \mod 3$.
This proves that among any 5 integers, sum of some 3 of them is divisible by 3.
Now, we have 11 integers. By the previous result, we can choose 3 of them such that there sum is divisible by 3. Denote this sum by $s_1$. Now, we are left with 8 integers, again, choose 3 of them such that there sum is divisible by 3. Denote this by $s_2$. Now, we are left with 5 integers. Choose $s_3$ similarly.
Thus we have 3 sums: $s_1, s_2, s_3$ (each of which are sums of 3 integers). These sums are divisible by 3. So, each of these sums are congruent to either 0 or 3 modulo 6.
Now, since there are 3 sums, and two residue clases ($, $), by Pigeonhole principle, one residue class must have two sums belonging to it. Let $s_i$ and $s_j$ be those sums. Either, $s_i \equiv s_j \equiv 0 \mod 6$ or $s_i \equiv s_j \equiv 3 \mod 6$. In both the cases, $s_i + s_j \equiv 0 \mod 6$.
Since, $s_i$ and $s_j$ are both sum of 3 integers, $s_i + s_j$ is a sum of 6 integers (which is divisible by 6). This completes the proof.