jmc

For integer $abcd$ to be divisible by $11$, then $a-b+c-d$ must be divisible by $11$. The only possibilities for $a-b+c-d$ are $-11$, $0$, and $11$. Possibility 1: $1-b+c-4=-11⟹c-b=-8$. This gives us two possible values: $c=0,b=8$, and $c=1,b=9$. Possibility 2: $1-b+c-4=0⟹c-b=3$. This gives us $7$ possible values, where $c$ is any integer from $3$ to $9$, and $b=c-3$. Possibility 3: $1-b+c-4=11⟹c-b=14$. This is impossible since the digit $c$ cannot be greater than $14$, Thus there are a total of $2+7=\boxed{9}$ possible values.