Let $S=2^3+3^4+5^4+7^4+\cdots+17497^4$ be the sum of the fourth powers of the first $2014$ prime numbers. Find the remainder when $S$ is divided by $240$.
1. **Using Fermat's Little Theorem**:
   Fermat's Little Theorem states that for any integer \(a\) and a prime \(p\), \(a^{p-1} \equiv 1 \pmod{p}\) if \(a\) is not divisible by \(p\). For the fourth power, we have:
   \[
   a^4 \equiv 1 \pmod{p} \quad \text{for primes } p > 3
   \]
   This means that for all primes \(p > 3\), \(p^4 \equiv 1 \pmod{p}\).

2. **Modulo 3**:
   - For \(p = 3\), \(3^4 \equiv 0 \pmod{3}\).
   - For all other primes \(p > 3\), \(p^4 \equiv 1 \pmod{3}\).
   - There are 2014 primes, and only one of them is 3.
   \[
   S \equiv 3^4 + 2013 \cdot 1 \equiv 0 + 2013 \equiv 2013 \equiv 0 \pmod{3}
   \]

3. **Modulo 5**:
   - For \(p = 5\), \(5^4 \equiv 0 \pmod{5}\).
   - For all other primes \(p \neq 5\), \(p^4 \equiv 1 \pmod{5}\).
   - There are 2014 primes, and only one of them is 5.
   \[
   S \equiv 5^4 + 2013 \cdot 1 \equiv 0 + 2013 \equiv 2013 \equiv 3 \pmod{5}
   \]

4. **Chinese Remainder Theorem**:
   - We have \(S \equiv 0 \pmod{3}\) and \(S \equiv 3 \pmod{5}\).
   - We need to solve the system:
     \[
     S \equiv 0 \pmod{3}
     \]
     \[
     S \equiv 3 \pmod{5}
     \]
   - Let \(S = 3k\). Then:
     \[
     3k \equiv 3 \pmod{5} \implies k \equiv 1 \pmod{5} \implies k = 5m + 1
     \]
     \[
     S = 3k = 3(5m + 1) = 15m + 3 \implies S \equiv 3 \pmod{15}
     \]

5. **Modulo 16**:
   - For any odd prime \(p\), \(p^4 \equiv 1 \pmod{16}\).
   - There are 2014 primes, all of which are odd.
   \[
   S \equiv 2014 \cdot 1 \equiv 2014 \equiv 14 \pmod{16}
   \]

6. **Combining results using the Chinese Remainder Theorem**:
   - We have:
     \[
     S \equiv 3 \pmod{15}
     \]
     \[
     S \equiv 14 \pmod{16}
     \]
   - Let \(S = 15k + 3\). Then:
     \[
     15k + 3 \equiv 14 \pmod{16} \implies 15k \equiv 11 \pmod{16} \implies k \equiv 11 \pmod{16}
     \]
     \[
     k = 16m + 11
     \]
     \[
     S = 15k + 3 = 15(16m + 11) + 3 = 240m + 165 + 3 = 240m + 168
     \]
     \[
     S \equiv 168 \pmod{240}
     \]

The final answer is \(\boxed{168}\).