The largest $n$ is $n=4$ if my calculations are ok, and I hope so. So, it's bashing, it boils down in the end to some casework, etc. The fact that numbers are integers is irrelevant, but they must be distinct. Take the biggest and the smallest number. By translating and scaling the assignments, we can assume that the biggest number is $1$ and the smallest is $0$. Suppose first, the corresponding edges of these two numbers are not incident. We take the tetrahedron of these edges and some casework yields that there is only one possibility (up to some symmetry) the numbers on that tetrahedron to form an arithmetic progressions, but some of the numbers must be equal ($0,1,1/3,2/3$ ).
Suppose now, $0$ and $1$ are assigned to incident edges, say, $0\to AB, 1\to AC$. In this case $1/2\to AC$. Consider any other vertex, say, $X$. Some calculation again gives only two possibilities
1) $XA=1/5, XB=2/5, XC=3/5$ and 2) $XA=5/7, XB=5/14, XC=3/7$.
This immediately implies $n\le 5$, since if there exist three vertices $X,Y,Z$, apart from $A,B,C$, some of assignments should be the same. If we assume $n=5$ with the vertices $A,B,C,X,Y$ then we must have $XA=1/5, XB=2/5, XC=3/5, YA=5/7, YB=5/14, YC=3/7$ up to interchanging $X$ and $Y$. An easy calculation shows that there is no number to be assigned to $XY$ in order to satisfy the requirements.