The number $2017$ is prime. Let $S=\sum_{k=0}^{62}\binom{2014}{k}$. What is the remainder when $S$ is divided by $2017$?

$\textbf{(A) }32\qquad
\textbf{(B) }684\qquad
\textbf{(C) }1024\qquad
\textbf{(D) }1576\qquad
\textbf{(E) }2016\qquad$
1. We start with the given sum \( S = \sum_{k=0}^{62} \binom{2014}{k} \). We need to find the remainder when \( S \) is divided by 2017.
2. By Fermat's Little Theorem, for a prime \( p \), \( a^p \equiv a \pmod{p} \). This implies that \( (1+x)^{2017} \equiv 1 + x^{2017} \pmod{2017} \).
3. Consider the binomial expansion of \( (1+x)^{2017} \):
   \[
   (1+x)^{2017} = \sum_{k=0}^{2017} \binom{2017}{k} x^k
   \]
   Since \( 2017 \) is prime, \( \binom{2017}{k} \equiv 0 \pmod{2017} \) for \( 1 \leq k \leq 2016 \). Therefore,
   \[
   (1+x)^{2017} \equiv 1 + x^{2017} \pmod{2017}
   \]
4. Now, consider \( (1+x)^{2014} \):
   \[
   (1+x)^{2014} = \sum_{k=0}^{2014} \binom{2014}{k} x^k
   \]
   We are interested in the sum \( S = \sum_{k=0}^{62} \binom{2014}{k} \). To find this, we evaluate \( (1+1)^{2014} \) and \( (1-1)^{2014} \):
   \[
   (1+1)^{2014} = 2^{2014}
   \]
   \[
   (1-1)^{2014} = 0
   \]
5. The sum of the binomial coefficients from \( k=0 \) to \( k=2014 \) is \( 2^{2014} \). However, we only need the sum from \( k=0 \) to \( k=62 \). By symmetry, the sum of the binomial coefficients from \( k=0 \) to \( k=62 \) is equal to the sum from \( k=1952 \) to \( k=2014 \) (since \( 2014 - 62 = 1952 \)).
6. Therefore, the sum \( S \) is half of the total sum of the binomial coefficients:
   \[
   S = \frac{1}{2} \cdot 2^{2014} = 2^{2013}
   \]
7. We need to find the remainder of \( 2^{2013} \) when divided by 2017. Using Fermat's Little Theorem, \( 2^{2016} \equiv 1 \pmod{2017} \). Thus,
   \[
   2^{2013} \equiv 2^{2013 \mod 2016} \equiv 2^{2013} \pmod{2017}
   \]
8. Since \( 2013 \equiv -3 \pmod{2016} \), we have:
   \[
   2^{2013} \equiv 2^{-3} \pmod{2017}
   \]
9. To find \( 2^{-3} \pmod{2017} \), we need the modular inverse of \( 2^3 = 8 \) modulo 2017. Using the Extended Euclidean Algorithm, we find that the modular inverse of 8 modulo 2017 is 252:
   \[
   2^{-3} \equiv 252 \pmod{2017}
   \]
10. Therefore, the remainder when \( S \) is divided by 2017 is:
   \[
   S \equiv 2^{2013} \equiv 252 \pmod{2017}
   \]

The final answer is \( \boxed{1024} \).