1. Define the problem and initial conditions:
We are given the function definitions:
$$[a] \text{ represents the largest integer less than or equal to } a,$$
$$\{a\} = a - [a].$$
We need to determine if there exist positive integers $m$ and $n$ and $n+1$ real numbers $x_0, x_1, \ldots, x_n$ such that:
$$x_0 = 428, \quad x_n = 1928,$$
and
$$\frac{x_{k+1}}{10} = \left[\frac{x_k}{10}\right] + m + \left\{\frac{x_k}{5}\right\}.$$

2. **Calculate the first few values of $x_i$:**
$$x_0 = 428.$$
Using the given recurrence relation:
$$\frac{x_1}{10} = \left[\frac{428}{10}\right] + m + \left\{\frac{428}{5}\right\}.$$
Calculate each term:
$$\left[\frac{428}{10}\right] = 42,$$
$$\left\{\frac{428}{5}\right\} = \frac{428}{5} - \left[\frac{428}{5}\right] = 85.6 - 85 = 0.6.$$
Therefore:
$$\frac{x_1}{10} = 42 + m + 0.6 = 42.6 + m,$$
$$x_1 = 10(42.6 + m) = 426 + 10m.$$

3. **Calculate $x_2$:**
$$\frac{x_2}{10} = \left[\frac{x_1}{10}\right] + m + \left\{\frac{x_1}{5}\right\}.$$
Since $x_1 = 426 + 10m$:
$$\left[\frac{426 + 10m}{10}\right] = 42 + m,$$
$$\left\{\frac{426 + 10m}{5}\right\} = \frac{426 + 10m}{5} - \left[\frac{426 + 10m}{5}\right] = 85.2 + 2m - (85 + 2m) = 0.2.$$
Therefore:
$$\frac{x_2}{10} = 42 + m + m + 0.2 = 42 + 2m + 0.2,$$
$$x_2 = 10(42 + 2m + 0.2) = 422 + 20m.$$

4. **Calculate $x_3$:**
$$\frac{x_3}{10} = \left[\frac{x_2}{10}\right] + m + \left\{\frac{x_2}{5}\right\}.$$
Since $x_2 = 422 + 20m$:
$$\left[\frac{422 + 20m}{10}\right] = 42 + 2m,$$
$$\left\{\frac{422 + 20m}{5}\right\} = \frac{422 + 20m}{5} - \left[\frac{422 + 20m}{5}\right] = 84.4 + 4m - (84 + 4m) = 0.4.$$
Therefore:
$$\frac{x_3}{10} = 42 + 2m + m + 0.4 = 42 + 3m + 0.4,$$
$$x_3 = 10(42 + 3m + 0.4) = 424 + 30m.$$

5. **Calculate $x_4$:**
$$\frac{x_4}{10} = \left[\frac{x_3}{10}\right] + m + \left\{\frac{x_3}{5}\right\}.$$
Since $x_3 = 424 + 30m$:
$$\left[\frac{424 + 30m}{10}\right] = 42 + 3m,$$
$$\left\{\frac{424 + 30m}{5}\right\} = \frac{424 + 30m}{5} - \left[\frac{424 + 30m}{5}\right] = 84.8 + 6m - (84 + 6m) = 0.8.$$
Therefore:
$$\frac{x_4}{10} = 42 + 3m + m + 0.8 = 42 + 4m + 0.8,$$
$$x_4 = 10(42 + 4m + 0.8) = 428 + 40m.$$

6. Analyze the periodic behavior:
Notice that:
$$x_1 \equiv 6 \pmod{10},$$
$$x_2 \equiv 2 \pmod{10},$$
$$x_3 \equiv 4 \pmod{10},$$
$$x_4 \equiv 8 \pmod{10}.$$
This pattern repeats every 4 steps. Therefore, for $x_n = 1928$, we must have:
$$x_{4k} = 1928.$$
Given:
$$x_{4k} = 428 + 40mk = 1928,$$
$$40mk = 1500.$$
Since 1500 is not divisible by 40, there are no positive integers $m$ and $k$ that satisfy this equation.

The final answer is False.