I have so far found two solutions (in both cases, ALICE is odd) so I'll post them here
$(A,L,I,C,E,B,O,N,F,U) = (2,0,3,6,5,4,9,1,8,7)$ or $(2,0,5,3,7,6,9,1,8,4)$
A five digit number minus a three-digit number results in a five-digit number with a different first digit
Hence, we must necessarily have
If as digits $E<B$ then the difference will carry over a $1$ and we'll necessarily get $C=U$, which is not allowed.
Hence $E>B$ and $U=C+1$ (also this means $E>N$)
The difference $C-O$ will necessarily carry a $1$ over to the hundreds place and since we need to carry over a one to the thousands place, we must have $I <B+1 \Rightarrow I < B$. Furthermore, we also must have $I<F$.
Since $A$, $E$ and $O$ are all greater than $N$, it follows that $N \le 6$.
As an example solution we can take $N=1$. Then, $A=2$, and the pairs $(B,E)$ and $(C,U)$ are consecutive digit pairs and together with $I$ and $F$ must constitute the entire set of numbers $(3,4,5,6,7,8)$. We just need to ensure $I-B-1$ gives $F (\bmod 10)$
So let's try $F=8$. Then we could have $I=3, B=4, E=5, C=6, U=7$ or perhaps $I=5, B=6, E=7, C=3, U=4$ which both seem to work. In fact, these are the only solutions in this case since $(I,B,E)$ must be consecutive as must $(C,U)$.
$F=7$ doesn't work since we need to assign two consecutive pairs to $(B,E)$ and $(C,U)$ from $(3,4,5,6)$ leaving $I=8$ which contravenes $I<B,F$
If $F=6$, then $I=3 \Rightarrow B=6=F$ (not allowed) $I=4$ means $C$ and $U$ cannot be consecutive and $I=5 \Rightarrow B=8$ and $E=9=O$ (not allowed).
It's not too hard to show that $F<5$ does not lead to any more solutions in this case ($N=1$).