The answer is $n=9$. Interpret the cards as elements of $\mathbb F_3^3$, and note that the matching condition of three cards with corresponding elements $a,b,c$ is equivalent to $a+b+c=0$. A construction for $n=9$ is $(1,1,0),(-1,1,0),(1,-1,0),(-1,-1,0),(0,0,1),(0,1,-1),(1,0,-1),(0,-1,-1),(-1,0,-1)$. Now, we show that among any 10 elements of $\mathbb F_3^3$ there are three distinct ones summing to 0.
Suppose we had a subset $A$ of $\mathbb F_3^3$ with $|A|=10$ and no three elements of $A$ summing to 0. First, we claim that there exist $w,x,y,z\in A$ with $w+x=y+z$. Otherwise, there cannot be any repeats if we take pairwise sums of the elements of $A$, a contradiction as $\binom{10}2>3^3$.
Now, take a linear transformation that sends $w,x,y,z$ to $(1,1,0),(-1,1,0),(-1,-1,0),(1,-1,0)$. The condition is clearly preserved under this transformation. Now, note that there cannot be any other points $(p,q,0)$ in $A$ and that for any $(a,b,c)$ in $A$ with $c\ne 0$, none of the points $(\pm 1-a,\pm 1-b,-c)$ can be in $A$. In particular, this means that if we have at least two points with $z$-coordinate $1$, then the set of available points with $z$-coordinate $-1$ either has size less than 3 or consists of three collinear points. This means that we cannot have at least two points in both the planes $z=1$ and $z=-1$. Thus, one of those planes must have 5 points. It is not hard to see that the answer to the 2-dimensional version of this question is 4 by some Pigeonhole, so we have a contradiction and are done.