We will prove that the answer of the problem is $9$.

Each card can be corresponded with a point in $\mathbb{Z}_3^3$. A line in $\mathbb{Z}_3^3$ is defined to be a subset of the form $\{P, P+V, P+2V\}$ where $P$ and $V$ are two elements in $\mathbb{Z}_3^3$, such that $V \neq 0$ (in $\mathbb{Z}_3^3$). It is easy to see that three elements of $\mathbb{Z}_3^3$ form a line, if and only if their sum (in $\mathbb{Z}_3^3$) is equal to $0$. For instance, in the following figure we can see three different lines. In this new language, the problem is to find the maximum number of points in $\mathbb{Z}_3^3$ such that no three of them are collinear.

The following lemma is the statement of the problem in $\mathbb{Z}_3^2$, and proof that the answer is not greater than $4$.

Lemma. There are no $5$ points in $\mathbb{Z}_3^2$ such that no three of them are collinear (definition of line in $\mathbb{Z}_3^2$ is similar).

Proof. Assume to the contrary that $S$ is a set in $\mathbb{Z}_3^2$ with more than $4$ elements such that no three points of $S$ are collinear. Let $P$ be one point in $\mathbb{Z}_3^2$. There are exactly $4$ lines (in $\mathbb{Z}_3^2$) passing through $P$. For example, if $P$ is a point in the corner, these are the lines containing $P$.

Since no three points of $S$ are collinear, each line passing through $P \in S$ contains at most one other point in this set. So there are at most five points in $S$ (Note that for any point in $\mathbb{Z}_3^2$ like $Q \neq P$, there is a unique line containing both $P$ and $Q$).
Now assume that $S$ has exactly $5$ elements. Consider $P \in S$. Again, since there are only four lines passing through $P$, and there are exactly four other points in $S$, every line passing through $P$ will contain another point of $S$.
It means that every line in $\mathbb{Z}_3^2$, intersects $S$ in exactly zero or two points. So there are $\binom{5}{2} = 10$ lines in $\mathbb{Z}_3^2$ containing two points of $S$. On the other hand, the total number of lines in $\mathbb{Z}_3^2$ is $12$ (for each of $9$ points in $\mathbb{Z}_3^2$, there are $4$ lines containing that point, and every line consists of $3$ different points). Therefore there are two lines $l_1$ and $l_2$ in $\mathbb{Z}_3^2$ having empty intersection with $S$. But the union of $l_1$ and $l_2$ contains at least $5$ points. This contradicts $|S| = 5$.

We call a subset $\mathcal{P}$ of $\mathbb{Z}_3^3$ a plane, if there exist $a, b, c, d \in \mathbb{Z}_3$ with at least one of $a, b$ and $c$ is non-zero (in $\mathbb{Z}_3$) and such that
$$\mathcal{P} = \mathcal{P}_d(a, b, c) = \{ (x, y, z) : ax + by + cz = d \pmod{3} \}.$$
It is easy to see that every plane consists of exactly $9$ points and for every three points $A, B$ and $C$ in $\mathbb{Z}_3^3$, either there exists a unique plane passing through them, or they are collinear. In the latter case there are exactly four different planes containing them.
We call two planes parallel, if they have no common points. Again, it is easy to check that if two planes $\mathcal{P}_1$ and $\mathcal{P}_2$ are parallel, there exist two different numbers $i, j \in \mathbb{Z}_3$ and $(a, b, c) \neq \vec{0}$ in $\mathbb{Z}_3^3$ such that
$$\mathcal{P}_1 = \mathcal{P}_i(a, b, c), \quad \mathcal{P}_2 = \mathcal{P}_j(a, b, c).$$
For any $(a, b, c) \neq \vec{0}$ (in $\mathbb{Z}_3^3$), all $27$ points in $\mathbb{Z}_3^3$ can be partitioned into three parallel planes $\mathcal{P}_0(a, b, c)$, $\mathcal{P}_1(a, b, c)$ and $\mathcal{P}_2(a, b, c)$. But since $\mathcal{P}_i(a, b, c) = \mathcal{P}_{-i}(-a, -b, -c)$, two points $(a, b, c)$ and $(-a, -b, -c)$ give the same partition. Therefore, there are exactly $\frac{26}{2} = 13$ ways to partition $\mathbb{Z}_3^3$ into three parallel planes. Note that if we consider all these $13$ partitions, every plane in $\mathbb{Z}_3^3$ will appear exactly in one of the partitions.
Now assume that there exist a set $S$ with $10$ elements satisfying the desired properties. Consider one of above partitions. According to the lemma, each plane in this partition contains at most $4$ points of $S$. Since there are $10$ points in total, there are two different type of partitions.
(1) Partitions such that two planes contain $4$ points of $S$, and the third one contains $2$.
(2) Partitions such that two planes contain $3$ points of $S$, and the third one contains $4$.
Let $m$ and $n$ be the total number of partitions of types (1) and (2), respectively. There are totally $13$ ways to partition the points. So we obtain the following equation.
$$m + n = 13.$$
We are going to find another equation for $m, n$.
For this reason, we count the number of elements of the following set in two different ways.
$$W = \{ (\{A, B\}, \mathcal{P}) : \mathcal{P} \text{ is a plane, } A, B \in \mathcal{P} \cap S \}$$
On one hand, if we count the number of pairs of points of $\mathcal{S}$ in each plane of partitions of type (1), we get a total of
$$\binom{4}{2} + \binom{4}{2} + \binom{2}{2} = 13$$
pairs of points for each partition of this type. And for the type (2), this number is
$$\binom{4}{2} + \binom{3}{2} + \binom{3}{2} = 12.$$
So the cardinality of $W$ is
$$13m + 12n.$$
On the other hand, for any two points $A, B \in \mathcal{S}$, the total amount of planes containing both $A$ and $B$ is $4$. Since there are totally $\binom{10}{2} = 45$ pairs, the cardinality of $W$ is $4 \times 45 = 180$, so we obtain $13m+12n = 180$. But this equation is clearly impossible when $m+n=13$ because
$$169 = 13m + 13n \geq 13m + 12n = 180,$$
Contradiction. So there are at most $9$ points with the desired properties. Also the following set of points is a valid example for $9$ points (look at the figure below).

Translating the solution into the main problem, the total number of cards we can choose is $9$. ■