Solution:

The diagram above shows that $n=2$ is possible (the path is $AHEFGHCDIHB$). Now suppose $n > 2$.

Note that if $X$ is any vertex except $A$ or $B$, then an even number of segments with endpoint $X$ must be in the path.

Let $F$ be the bottom left-hand vertex. Two sides of the triangle $EFG$ are in the path, so at least one of $EF$ and $FG$ is. But $EF$ and $EG$ are the only segments with endpoint $F$, so an even number of them must be in the path, so both are in the path. Hence, again considering $EFG$, $EG$ is not in the path. Hence, considering $EHG$, $EH$ and $HG$ are in the path.

$E$ has an even number of segments on the path, so $CE$ is not on the path. Hence (considering $CEH$) $CH$ is on the path. Similarly, $GJ$ is not on the path and $HJ$ is on the path. An even number of segments at $H$ are on the path, so $DH$ and $HI$ are either both on the path or neither is on the path. But (considering $DHI$) at least one must be, so they both are. Hence $DI$ is not, and $CD$ is not.

Since $n>2$, $C$ is not the top left vertex. Considering $MCD$, $MC$ and $MD$ are both on the path. Considering $DLI$, $DL$ is on the path. There must be an even number of segments at $D$, so $DP$ is on the path. Hence $MP$ is not. Now $M$ cannot be the top left vertex (with $n=3$) because then it should have an odd number of segments, whereas it would have two ($MC$ and $MD$). So there must be a vertex $N$ above $M$. Considering $NMP$, $MN$ must be in the path. But now $M$ has an odd number of segments. Contradiction.