Given a positive integer $n$, we aim to find all $n$-tuples of real numbers $(x_1, x_2, \ldots, x_n)$ such that
$$f(x_1, x_2, \cdots, x_n) = \sum_{k_1=0}^{2} \sum_{k_2=0}^{2} \cdots \sum_{k_n=0}^{2} \left| k_1 x_1 + k_2 x_2 + \cdots + k_n x_n - 1 \right|$$
attains its minimum.

To solve this, we first claim that the minimum is achieved when all $x_i$ are equal. Specifically, we seek to show that the minimum occurs at $x_1 = x_2 = \cdots = x_n = \frac{1}{n+1}$.

### Proof:
1. Symmetry Argument:
Let $y = \frac{x_1 + x_2}{2}$. We will prove that
$$f(x_1, x_2, x_3, \cdots, x_n) \geq f(y, y, x_3, \cdots, x_n).$$
For any $k, m, c \in \mathbb{R}$,
$$|kx_1 + mx_2 + c| + |mx_1 + kx_2 + c| \geq |(k+m)(x_1 + x_2) + 2c| = |ky + my + c| + |my + ky + c|$$
by the triangle inequality.

2. Application of Inequality:
Applying the above inequality, let $c = \sum_{j=3}^n k_j x_j - 1$ and summing over all $c$ for all $(k_3, \cdots, k_n) \in \{0, 1, 2\}^{n-2}$ where $k_1 \neq k_2$ gives the desired result.

3. Reduction to Single Variable:
Now, let $x_1 = x_2 = \cdots = x_n = x$, and we need to minimize
$$g(x) = f(x, x, x, \cdots, x).$$
Let $H(n, t)$ be the number of solutions to $k_1 + \cdots + k_n = t$ where $k_j \in \{0, 1, 2\}$ for $j = 1, \cdots, n$. Observe that $H(n, t) \leq H(n, n)$ and $H(n, t) = H(n, 2n - t)$.

4. Minimization:
Therefore,
$$g(x) = \sum_{j=0}^{2n} H(n, j) |jx - 1| = \sum_{j=0}^{2n} j H(n, j) \left| x - \frac{1}{j} \right|.$$
We claim that $g(x)$ is minimized at $x = \frac{1}{n+1}$.

5. Uniqueness:
It remains to show that $x_1 = x_2$ is forced. In the smoothing process, any move that produces a smaller sum is not possible. If $x_1 = x_2$, then there exist $k_1 \neq k_2$ such that $k_1 x_1 + \cdots + k_n x_n = 1$. Now, let $y_1, y_2$ such that $y_1 + y_2 = 2x_1$. Then,
$$|k_1 y_1 + k_2 y_2 + k_3 x_3 + \cdots + k_n x_n - 1| + |k_2 y_1 + k_1 y_2 + k_3 x_3 + \cdots + k_n x_n - 1| > 2 |k_1 x_1 + k_2 x_2 + k_3 x_3 + \cdots + k_n x_n - 1|,$$
so the $x_i$'s must be constant.

Thus, the minimum value of the function is attained when $x_1 = x_2 = \cdots = x_n = \frac{1}{n+1}$.

The answer is: $\boxed{\left( \frac{1}{n+1}, \frac{1}{n+1}, \ldots, \frac{1}{n+1} \right)}$.