By Lucas theorem, if $n=2017+t$ and $k'=k\pmod{2017}$, then $\binom{n}{k}$ is congruent to either $\binom{1}{0}\binom{t}{k'}=\binom{t}{k'}$ or $\binom{1}{1}\binom{t}{k'}=\binom{t}{k'}$. If $k'>t$ we clearly have $\binom{t}{k'}=0$. Also, note that in our problem $\binom{t}{k'}\leq \binom{13}{6}$ which is clearly less than $2017$. Therefore, our sum is $$\sum_{t=0}^{13}(-1+2\sum_{k'=0}^t \frac{1}{2017}\binom{t}{k'}=\frac{1}{2017}\sum_{t=0}^{13} (-1+2\cdot 2^t)=\frac{-13+2(2^{14}-1)}{2017}$$