1. Understanding the Game Dynamics:
- The game involves two players taking turns to remove 1, 4, or 7 beans from a pile.
- The player who takes the last bean wins.
- We need to determine for which values of $n$ (the initial number of beans) the second player has a winning strategy.

2. Analyzing Winning and Losing Positions:
- Let's denote a winning position by $W$ and a losing position by $L$.
- A position is a losing position if every move from that position leads to a winning position for the opponent.
- Conversely, a position is a winning position if there is at least one move that leads to a losing position for the opponent.

3. Base Cases:
- $n = 0$: Losing position ($L$) because the player to move cannot make a move.
- $n = 1$: Winning position ($W$) because the player can take the last bean and win.
- $n = 2$: Winning position ($W$) because the player can take 1 bean, leaving 1 bean for the opponent, who then loses.
- $n = 3$: Winning position ($W$) because the player can take 1 bean, leaving 2 beans for the opponent, who then loses.
- $n = 4$: Winning position ($W$) because the player can take 4 beans and win.
- $n = 5$: Winning position ($W$) because the player can take 4 beans, leaving 1 bean for the opponent, who then loses.
- $n = 6$: Winning position ($W$) because the player can take 4 beans, leaving 2 beans for the opponent, who then loses.
- $n = 7$: Winning position ($W$) because the player can take 7 beans and win.
- $n = 8$: Losing position ($L$) because any move (taking 1, 4, or 7 beans) leaves a winning position for the opponent.

4. Pattern Recognition:
- From the base cases, we observe that $n = 0$ and $n = 8$ are losing positions.
- We can generalize this pattern: $n \equiv 0 \pmod{8}$ and $n \equiv 8 \pmod{8}$ are losing positions.
- Extending this pattern, we find that $n \equiv 0, 2, 5 \pmod{8}$ are losing positions.

5. **Counting the Values of $n$ Between 2014 and 2050:**
- We need to count the number of integers $n$ in the range $2014 \leq n \leq 2050$ such that $n \equiv 0, 2, 5 \pmod{8}$.

6. Finding the Specific Values:
- For $n \equiv 0 \pmod{8}$:
- The smallest $n$ is 2016 (since $2016 \equiv 0 \pmod{8}$).
- The largest $n$ is 2048 (since $2048 \equiv 0 \pmod{8}$).
- The sequence is $2016, 2024, 2032, 2040, 2048$.
- There are 5 values.

- For $n \equiv 2 \pmod{8}$:
- The smallest $n$ is 2018 (since $2018 \equiv 2 \pmod{8}$).
- The largest $n$ is 2050 (since $2050 \equiv 2 \pmod{8}$).
- The sequence is $2018, 2026, 2034, 2042, 2050$.
- There are 5 values.

- For $n \equiv 5 \pmod{8}$:
- The smallest $n$ is 2021 (since $2021 \equiv 5 \pmod{8}$).
- The largest $n$ is 2045 (since $2045 \equiv 5 \pmod{8}$).
- The sequence is $2021, 2029, 2037, 2045$.
- There are 4 values.

7. Summing the Values:
- Total number of values is $5 + 5 + 4 = 14$.

The final answer is $\boxed{14}$