Let $F_1, F_2, F_3 \cdots$ be the Fibonacci sequence, the sequence of positive integers satisfying $$F_1 =F_2=1$$ and $$F_{n+2} = F_{n+1} + F_n$$ for all $n \ge 1$.

Does there exist an $n \ge 1$ such that $F_n$ is divisible by $2014$? Prove your answer.
To determine if there exists an \( n \ge 1 \) such that \( F_n \) is divisible by \( 2014 \), we will use the Pigeonhole Principle and properties of the Fibonacci sequence modulo \( 2014 \).

1. **Fibonacci Sequence Modulo \( 2014 \)**:
   The Fibonacci sequence is defined as:
   \[
   F_1 = 1, \quad F_2 = 1, \quad F_{n+2} = F_{n+1} + F_n \quad \text{for all } n \ge 1.
   \]
   We consider the sequence modulo \( 2014 \):
   \[
   F_1 \mod 2014, \quad F_2 \mod 2014, \quad F_3 \mod 2014, \quad \ldots
   \]
   Since there are only \( 2014 \times 2014 = 4056196 \) possible pairs \((F_n \mod 2014, F_{n+1} \mod 2014)\), by the Pigeonhole Principle, the sequence must eventually repeat.

2. **Periodicity**:
   Let \( m \) be the smallest positive integer such that the sequence \( F_n \mod 2014 \) is periodic with period \( m \). This means:
   \[
   F_{n+m} \equiv F_n \mod 2014 \quad \text{for all } n.
   \]
   Specifically, we have:
   \[
   F_{m+1} \equiv F_1 \equiv 1 \mod 2014 \quad \text{and} \quad F_{m+2} \equiv F_2 \equiv 1 \mod 2014.
   \]

3. **Existence of \( n \) such that \( F_n \) is divisible by \( 2014 \)**:
   Since the sequence is periodic, we can write:
   \[
   F_{m+1} \equiv 1 \mod 2014 \quad \text{and} \quad F_{m+2} \equiv 1 \mod 2014.
   \]
   This implies:
   \[
   F_{m+2} = F_{m+1} + F_m \equiv 1 + F_m \mod 2014.
   \]
   Therefore:
   \[
   1 \equiv 1 + F_m \mod 2014 \implies F_m \equiv 0 \mod 2014.
   \]
   Hence, \( F_m \) is divisible by \( 2014 \).

Conclusion:
By the periodicity of the Fibonacci sequence modulo \( 2014 \), we have shown that there exists an \( n \ge 1 \) such that \( F_n \) is divisible by \( 2014 \).

The final answer is \( \boxed{ n } \) such that \( F_n \) is divisible by \( 2014 \).