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\neq 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} $.