Problem 1442

1. Label the Board:
We label the $16$ cells on the $4 \times 4$ board using coordinates $(x, y)$ where $x, y \in \{1, 2, 3, 4\}$.

2. Assign Complex Numbers:
We assign the number $\omega^{i+j}$ to the cell $(i, j)$ where $\omega = e^{\frac{i\pi}{3}}$. Note that $\omega$ is a primitive 6th root of unity, so $\omega^6 = 1$.

3. Initial Sum Calculation:
Initially, the sum of the numbers on the board is $k_1\omega + k_2\omega^2 + k_3$ where $(k_1, k_2, k_3) \in \{(5, 5, 5), (4, 6, 5), (5, 6, 4)\}$. This is because there are 15 stones on the board, and the sum of the exponents $i+j$ modulo 3 will distribute among $\omega, \omega^2$, and $1$.

4. Sum Change During Moves:
During each move, the sum of the numbers on the board changes. If a stone at $(i, j)$ jumps over a stone at $(i, j+1)$ to an empty cell at $(i, j+2)$, the sum changes from $a_1\omega + a_2\omega^2 + a_3$ to $a_1'\omega + a_2'\omega^2 + a_3'$ where $(a_1', a_2', a_3') \in \{(a_1+1, a_2-1, a_3-1), (a_1-1, a_2+1, a_3-1), (a_1-1, a_2-1, a_3+1)\}$.

5. Parity Consideration:
The parity of $|a_1 - a_2|$ and $|a_2 - a_3|$ remains unchanged during each move. This is because each move involves a change in the coefficients of $\omega, \omega^2$, and $1$ that preserves the parity of their differences.

6. Initial Parity:
For the initial configuration $(k_1, k_2, k_3)$, we have $|5 - 5| = 0$, $|4 - 6| = 2$, and $|5 - 6| = 1$. Therefore, the parity of $|a_1 - a_2|$ and $|a_2 - a_3|$ is fixed.

7. Final Sum:
To end up with exactly one stone on the board, the sum of the numbers must be $\omega$, $\omega^2$, or $1$. However, since the initial parity is $0$, the final sum cannot be $\omega$, $\omega^2$, or $1$.

8. Empty Cell Position:
Therefore, the empty cell cannot have coordinates $(i, j)$ where $i + j \equiv 2 \pmod{3}$. This leaves the 8 border non-corner cells and the 2 center cells.

9. Symmetry Consideration:
By symmetry, we can eliminate the two center cells, leaving us with 8 border non-corner cells as possible positions for the empty cell.

10. Verification:
It is possible to show that starting with the empty cell in one of these 8 positions, we can end up with exactly one stone on the board through a series of valid moves.

The final answer is the 8 border non-corner cells.