I hope I have not messed up stupid details. Answer: $9$ It is easier to start with finding an example for a reasonably small $n$. Suppose the first three matches are wins and play $(1, 4)$, $(2, 5)$, $(3, 6)$, so ranking after them is $1 2 3 4 5 6 | 3 3 3 0 0 0$. For the next three, make $(3,5)$ and $(2,4)$ draws and $(1,6)$ win for $1$, to get $6 4 4 1 1 0$. Now let $7$-th be a $(1,2)$ win for $2$, let $8$-th be a $(3, 4)$ draw and $9$-th be a $(5,6)$ win for $6$, to get $6 7 5 2 1 3$. Now we prove $n\leq 8$ is impossible. The results being distinct and non-zero imply their sum is at least $6 + 5 + 4 + 3 + 2 + 1 = 21$ and hence $n\geq 7$; but equality to hold requires there to be no draws (as a draw contributes a total of $2$ and not $3$ points) and at the same time to have $6$, $5$, $4$, $3$, $2$, $1$ as points, contradiction. Finally, suppose $n=8$. If $x$ is the total number of draws (so $8-x$ are win-losses), the total sum of scores is $2x + 3(8-x) = 24 - x$. Since the total is at least $21$, we have $x\leq 3$. If $x=0$, then all scores are multiples of $3$ and their sum is at least $3 + 6 + 9 + 12 + 15 + 18 > 24$, nope. If $x=1$, the only options of distinct non-zero scores are $7 6 4 3 2 1$ and $8 5 4 3 2 1$, but the $2$ comes from two draws, contradiction. If $x=2$, the only option of non-zero scores is $7 5 4 3 2 1$, but then $5$ and $2$ need at least three distinct matches of draws to have two draws in their score (as they both cannot play twice), contradiction. Hmm, but actually $x=3$ is possible... and I think the next post is the correct solution, but leaving this one for the moral.

Real answer: $8$. Suppose the first three matches are wins and play $(1, 4)$, $(2, 5)$, $(3, 6)$, so ranking after them is $1 2 3 4 5 6 | 3 3 3 0 0 0$. For the next three, make $(1,6)$ a win for $1$, let $(2,4)$ and $(3,5)$ be draws, to get $6 4 4 1 1 0$. Finally, a $(3,4)$ draw and a $(5,6)$ win for $6$ leads to $6 4 5 2 1 3$. The results being distinct and non-zero imply their sum is at least $6 + 5 + 4 + 3 + 2 + 1 = 21$ and hence $n\geq 7$; but equality to hold requires there to be no draws (as a draw contributes a total of $2$ and not $3$ points) and at the same time to have $6$, $5$, $4$, $3$, $2$, $1$ as points, contradiction.