Seems like a paper-based problem... Let us treat integer tuples as vectors on $\mathbb{Z}_3^3$, and integers on $\mathbb{Z}_3$. Since no vector can appear three times, we have at least $10$ distinct vectors on $\mathbb{Z}_3^3$. Let $v_1, v_2, \cdots , v_{10}$ be the given ten vectors, and suppose that there are no subscripts $1\le i<j<k \le 10$ satisfying $v_i+v_j+v_k=0$. Rather than dealing with $3$ entries, let us consider the 2D problem first: Claim. Let $W=\{ w_1, \cdots , w_t \}$ be pairwise distinct vectors on $\mathbb{Z}_3^2$. If there are no subscripts $i<j<k$ such that $w_i+w_j+w_k=0$, then $t \le 4$. Proof of Claim. Consider a parallel transformation of the vectors $w_i$. Note that the sum $w_i+w_j+w_k$ is invariant under the transformation, henceforth we can WLOG assume that $w_1=0$. Note that we can pair the vectors of $\mathbb{Z}_3^2-\{(0,0)\}$ so that every pair sum up to $0$. Since there are four pairs, and there are at most one $w_i$'s among each pair, so $t \le 5$. Now suppose that $t=5$. Note that there exist a residue class $\bar{a}$ modulo $3$ so that there are at most one vector with first entry $a$. Take a parallel transformation about this vector, and we get a figure with ${(0,0)} \in W$, but $(0,1), (0,2) \not \in W$. Now consider the above pairing again; this leads to $t \le 4$. (to visualize this, consider an $3 \times 3$ grid, and color all the squares corresponding to the vectors, and see what happens!) Hence the proof is complete. $\square$ Now, for $i \in \mathbb{Z}_3$, denote $V_i$ by the set of vector $v_j$s with first entry $i$. If $|V_i| \ge 5$ for some $i$, then by the claim, there are three vectors of $V_i$ so that their projection on the $yz$-plane sum up to $(0,0)$. Since their first entry are all same, that three vectors are the desired vectors. Suppose otherwise, so that $|V_i| \le 4$ for every residue class $i$. Note that any parallel transformation, and any scalar multiplication(in $\mathbb{Z}_3^{*}$) doesn't change the existence of "three vectors that sum up to the zero vector", so WLOG assume that $|V_0| \ge |V_1| \ge |V_2|$. Now, the remaining cases are $\left( |V_0|, |V_1|, |V_2| \right)=(4,4,2), (4,3,3)$. Case 1: $\left( |V_0|, |V_1|, |V_2| \right)=(4,4,2)$. Project all vectors on the $yz$-plane(so that their first entry dies out). Let $W_0, W_1, W_2$ be the corresponding vectors of $V_0, V_1, V_2$ under projection. Let $B_2=\{ a, b \}$, and consider the sets $W_{11} \equiv -a-W_1$, $W_{12} \equiv -b-W_1$. Since no vectors $p \in V_0, q \in V_1, r \in V_2$ sum up to the zero vector, so both sets $W_{11}$, $W_{12}$ are disjoint with $W_0$. But $\mathbb{Z}_3^2$ has only $9$ elements, and $|W_0|=|W_{11}|=|W_{22}|=4$, so $|W_{11} \cap W_{22}| \ge 3$. For convinence, let $W_1=\{a,b,c,d\}$, and let us construct a bipartite graph $G$ with vertices $\{a,b,c,d \} \cup \{ a',b',c',d'\}$, and $\{ \square, \triangle' \}$ is an edge if and only if $-a-\square=-b-\triangle'$. Note that the set $\{ \square, \square'\}$ is always independent, and every vertex has degree $\le 1$. Since there are at least $3$ edges, we can find some distinct vectors $\triangle, \square, \hexagon$ such that both $\{ \triangle, \square'\}$ and $\{\square ,\hexagon'\}$ are edges on $G$. Then $\triangle -\square=\square -\hexagon$, so $\triangle+\square+\hexagon=3\square =0$, which is exactly what we was looking for. Case 2: $\left( |V_0|, |V_1|, |V_2| \right)=(4,3,3)$. This is similar with the above case, we construct three transformations of $W_1$: $W_{11}, W_{12}, W_{13}$. By constructing graphs, we get $|W_{1i} \cap W_{1j}| \le 1$ for any distinct subscripts $i,j$. Now using PIE, an similar "counting elements of $\mathbb{Z}_3^2$" argument works. Hence we are done. $\blacksquare$

This is an well-known problem related to the EGZ theorem. $f(n,d)$ : smallest number of lattice points in $d$-dimensional euclidean space to guarantee the existence of $n$ points with lattice point centroid. For $d=2$, it's the Kemnitz's conjecture // link : https://en.wikipedia.org/wiki/Kemnitz%27s_conjecture Problem : Prove $f(3,3) \ge 19$ Actually, $f(3,3)=19$. It was solved by German mathematician 'Heiko Harborth' at 1973. // link : https://eudml.org/doc/151373 Idea for this problem

lminsl wrote: both sets $W_{11}$, $W_{12}$ are coprime with $W_0$ I don't understand this concept, "$A$ coprime $B$" with $A,B$ be sets.

analysis90 wrote: lminsl wrote: both sets $W_{11}$, $W_{12}$ are coprime with $W_0$ I don't understand this concept, "$A$ coprime $B$" with $A,B$ be sets. Sorry for the confusion Fixed!

lminsl wrote: For convinence, let $W_1=\{a,b,c,d\}$, and let us construct a bipartite graph $G$ with vertices $\{a,b,c,d \} \cup \{ a',b',c',d'\}$, and $\{ \square, \triangle' \}$ is an edge if and only if $-a-\square=-b-\triangle'$. Note that the set $\{ \square, \square'\}$ is always independent, and every vertex has degree $\le 1$. Since there are at least $3$ edges, we can find some distinct vectors $\triangle, \square, \hexagon$ such that both $\{ \triangle, \square'\}$ and $\{\square ,\hexagon'\}$ are edges on $G$. Then $\triangle -\square=\square -\hexagon$, so $\triangle+\square+\hexagon=3\square =0$, which is exactly what we was looking for. I tried to read this text, but I also didn't understand it. What is $a',b',c',d'$? $\square, \square',\triangle,\triangle',\dots$ Can you explain it clearly?

I'll point out a few things I don't understand with Iminsl's solution. First of all, I think that instead of $W_{22}$, he (or she) means to say $W_{12}$. Second, I think he meant to say $B_2=\{a,b\}$ and $B_2 \in W_2$. Also, I don't understand what $\{a',b',c',d'\}$ are as Iminsl never defined these terms. Is the set the same as $\{a,b,c,d\}$?