Answer: C

Solution:

Working modulo $2017,$ the identity $\binom{2014}{k} \cdot k! \cdot (2014-k)! = 2014!$ together with $2016 \cdot 2015 \cdots (2015-k)$ $\equiv (-1)^k (k+2)!$ leads to $$\begin{gathered} 2\binom{2014}{k} \equiv (-1)^k \\ {}\cdot (k+2)(k+1) \pmod{2017}, \end{gathered}$$ so $\binom{2014}{k} \equiv (-1)^k \binom{k+2}{2}.$

Then $$\begin{gathered} S \equiv \sum_{k=0}^{62} (-1)^k \binom{k+2}{2} \\ = 1 \\ {}+ \sum_{k=1}^{31}\left[\binom{2k+2}{2} - \binom{2k+1}{2}\right] \\ = 1 + \sum_{k=1}^{31}(2k+1). \end{gathered}$$

The remaining sum is $3 + 5 + \cdots + 63 = 1023,$ so $S \equiv 1 + 1023$ $= 1024 \pmod{2017}.$

Thus, the correct answer is C.