I am pretty sure this is not an intended solution, I just want to point out that the way the questions is posed is not completely satisfying.
There are a lot of solutions which separate the sets. Let us consider just a simple mapping $f:C \rightarrow \mathbb{Z}_{36}$, where each word is an element in $C^6$ and $(c_1,...c_6) \in C^6$ is evaluated as $\sum f(c_i) \bmod 36$. Here is such a mapping
{'1': 10, '0': 32, '3': 5, '2': 35, '5': 25, '4': 13, '7': 26, '6': 23, '9': 2, '8': 18, 'A': 4, 'C': 3, 'B': 1, 'E': 34, 'D': 20, 'G': 6, 'F': 0, 'I': 15, 'H': 8, 'K': 30, 'J': 2, 'M': 21, 'L': 9, 'O': 29, 'N': 16, 'Q': 17, 'P': 7, 'S': 14, 'R': 11, 'U': 19, 'T': 0, 'W': 12, 'V': 33, 'Y': 31, 'X': 28, 'Z': 17}
which produces an element-wise sum congruent $0 \bmod 36$.
43RM26 0
7330GE 0
MOE21I 0
GB1XUH 0
XMATFU 0
91A6JY 0
KA40A5 0
WMYMDC 0
CXSOZQ 0
.........
EPWC4X 25
YK7JSA 35
NOS0QZ 17
GSIG0C 4
S5QH80 6
9W2JVM 33
JN8J2H 9
STVFSS 3
PXGSG7 15
Another is
{'1': 24, '0': 3, '3': 17, '2': 4, '5': 8, '4': 29, '7': 2, '6': 25, '9': 18, '8': 9, 'A': 30, 'C': 13, 'B': 0, 'E': 19, 'D': 33, 'G': 14, 'F': 1, 'I': 16, 'H': 15, 'K': 8, 'J': 6, 'M': 23, 'L': 31, 'O': 22, 'N': 26, 'Q': 20, 'P': 28, 'S': 7, 'R': 10, 'U': 21, 'T': 35, 'W': 11, 'V': 32, 'Y': 5, 'X': 34, 'Z': 12}
This fits with the hint that a randomly chosen string is valid with probability $1/36$.
It is easy to find since most of the strings in the valid set have one or more free variables which can be used to zero out the sum. Just pick a random map for the variables that occur at least twice. Then, pick the remaining ones such that it does not occur in the set (I am assuming here that the map is invertible -- if it is not, the problem is even easier). This requires little bit of iteration. Then check if the invalid strings are all non-zero. Otherwise, repeat.
