jmc

Since $T$ must be divisible by $14$, it must be divisible by $2$ and $7$. Since it's divisible by $2$, the last digit must be even, so the units digit must be $0$. $T$ must also be divisible by $7$. Let $R$ be the number obtained by taking $T$ and chopping off the last digit, $0$. In order for $T$ to be divisible by $7$, $R$ must be divisible by $7$, and $R$ must also be made up of $1$'s and $0$'s. If $R$ has one digit, it must be $1$ (since $T\ne 0$), which isn't divisible by $7$. If $R$ has $2$ digits, it must be $10$ or $11$, neither of which are divisible by $7$. If $R$ has $3$ digits, it must be $100$, $101$, $110$, or $111$. Here we can use the divisibility rule for $7$, by chopping off the last digit, multiplying it by two, and subtracting it from the rest, to see that none of these values are divisible by $7$ either. If $R$ has $4$ digits, we can check as we go along: if $R=1000$, then the divisibility rule reduces our checking to whether $100$ is divisible by $7$, and we already know it's not. If $R=1001$, then the divisibility rule asks if $98$ is divisible by $7$--and it is! So $R=1001$ works. This means $T=10010$. We want the quotient $\frac{10010}{14}=\boxed{715}$.