Prove that, among 9 tuples of integers $(a_i, b_i)$, there exists 3 tuples $(a_i,b_i), (a_j,b_j), (a_k,b_k)$ such that $3\mid (a_i+a_j+a_k)$ and $3 \mid (b_i+b_j+b_k)$.
So under modulo 3 there would be 9 distinct cases, if all distinct cases occur, so we take $(0,0), (1,1), (2,2)$. We also note that either $a_i$ are same for all $i$ or they are all distinct, and same goes for the $b_i$. Thinking in the direction of the pigeonhole principle, but still not able to prove it.
There are several cases and sub-cases to consider.
Case $1$: Among the first set of values, i.e., the $a_n$, there are at least $5$ which have the same remainder modulo $3$.
As listed in this answer, two posts here which prove that, among any $5$ integers, there are always $3$ of them which sum to a multiple of $3$ are How do I prove that among any $5$ integers, you are able to find $3$ such that their sum is divisible by $3$? and Prove that for any set of 5 integers, there is at least one subset of 3 integers whose sum is divisible by 3. Thus, we can find $3$ values among the corresponding $5$ $b_n$ values which sum to a multiple of $3$. If the common $a_n$ remainder is $m$, then $a_i + a_j + a_k \equiv 3m \equiv 0\pmod{3}$ regardless of which indices are used for the $3$ $b_n$, so we can use those indices.
Case $2$: There are at most $4$ values among the $a_n$ with any one remainder.
By the pigeonhole principle, we must have at least one $a_n$ with each congruence (as otherwise we have at most $4$ each of $2$ different congruences). Also, there are at least $3$ $a_n$ values of $2$ different congruences (since we either have $3$ of each one, or one has $4$, but the other $2$ different congruences can't both have at most $2$ representatives), with these sets of tuples with the $a_i$ values being called $g_1$ and $g_2$, and the remaining set with the third congruence called $s_1$. Consider the following sub-cases regarding the $b_n$ congruences.
Case $2A$: Among either $g_1$ or $g_2$, there's at least $3$ with the same $b_n$ congruence or all $3$ of the possible congruences are represented.
We can just choose those $3$ indices to get the requested result.
Case $2B$: There are $2$ different congruences represented among the $b_n$ of each of $g_1$ and $g_2$.
Note that by choosing one tuple from each of the sets $g_1$, $g_2$ and $s_1$, we get that $a_i + a_j + a_k \equiv 0 + 1 + 2 \equiv 0 \pmod{3}$.
Case $2B$-$1$: These $2$ sets of congruences among $g_1$ and $g_2$ are both the same.
If the $b_n$ value from an $s_1$ member has one of those $2$ congruences, we can use that $s_1$ value along with one of each of these from $g_1$ and $g_2$. Otherwise, if the $b_n$ value is not among those $2$ congruences, we can use the $s_1$ value along with one of the congruences from $g_1$ and the other congruence from $g_2$.
Case $2B$-$2$: These $2$ sets of congruences among $g_1$ and $g_2$ are not the same.
There's one congruence overlap between them, and each have a different other congruence. If the $b_n$ value from $s_1$ has the overlap congruence value, then we can choose that among each of $g_1$ and $g_2$. Otherwise, from either $g_1$ or $g_2$ which has this $b_n$ congruence of the $s_1$ member value, choose the overlap congruence from this set, and the third congruence from the other set, with this working since the $b_n$ congruences would then all be different.
Since this covers all of the possibilities, it proves what was requested.
