A domino is a $ 1 \times 2 $ or $ 2 \times 1 $ tile.
Let $n \ge 3 $ be an integer. Dominoes are placed on an $n \times n$ board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. The value of a row or column is the number of dominoes that cover at least one cell of this row or column. The configuration is called balanced if there exists some $k \ge 1 $ such that each row and each column has a value of $k$. Prove that a balanced configuration exists for every $n \ge 3 $, and find the minimum number of dominoes needed in such a configuration.
1. **Define the problem and notation:**
   - A domino is a $1 \times 2$ or $2 \times 1$ tile.
   - We need to place dominoes on an $n \times n$ board such that each row and each column has the same value $k$, where the value of a row or column is the number of dominoes that cover at least one cell of this row or column.
   - We need to prove that such a balanced configuration exists for every $n \ge 3$ and find the minimum number of dominoes needed.

2. **Set up the equations:**
   - Let $r$ be the number of row-aligned dominoes and $c$ be the number of column-aligned dominoes.
   - Each row-aligned domino covers 1 row and 2 columns.
   - Each column-aligned domino covers 2 rows and 1 column.
   - Each row is covered $k$ times, so the total number of row coverages is $r + 2c$.
   - Each column is covered $k$ times, so the total number of column coverages is $2r + c$.
   - Therefore, we have the equations:
     \[
     \frac{r + 2c}{k} = n \quad \text{(1)}
     \]
     \[
     \frac{2r + c}{k} = n \quad \text{(2)}
     \]

3. **Sum the equations:**
   - Adding equations (1) and (2):
     \[
     \frac{r + 2c + 2r + c}{k} = 2n
     \]
     \[
     \frac{3r + 3c}{k} = 2n
     \]
     \[
     r + c = \frac{2nk}{3} \quad \text{(3)}
     \]

4. **Analyze the divisibility:**
   - If $3 \mid n$, then $\frac{2nk}{3}$ is an integer for $k = 1$.
   - If $3 \nmid n$, then $\frac{2nk}{3}$ is first an integer for $k = 3$.

5. **Construct the configurations:**
   - For $n \equiv 0 \pmod{3}$:
     - The minimal $k = 1$ is feasible.
     - For $n = 3$, we can place dominoes as follows:
       \[
       \begin{array}{ccc}
       (2,1) & (3,1) & \\
       (1,2) & (1,3) & \\
       \end{array}
       \]
     - For larger $n$, we can place $n = 3$ grids along the diagonal.

   - For $n \not\equiv 0 \pmod{3}$:
     - The minimal $k = 3$ is feasible.
     - For $n = 4, 5, 6, 7$, we can construct configurations with $k = 3$.
     - For larger $n$, we can successively add $n = 4$ grids along the diagonal.

6. **Determine the minimum number of dominoes:**
   - For $n \equiv 0 \pmod{3}$, the minimum number of dominoes is $\frac{2n}{3}$.
   - For $n \not\equiv 0 \pmod{3}$, the minimum number of dominoes is $2n$.

The final answer is $\boxed{\frac{2n}{3} \text{ if } 3 \mid n;~2n \text{ otherwise}}$.