Let's express the fraction $\frac{674}{385}$ as the sum of three fractions such that the sum of the numerators is equal to the sum of the digits in the three denominators.
Let's reasonably interpret the problem such that the three sought fractions are positive and irreducible, and their sum is the fraction $\frac{674}{385}$, not any of its extended forms.

Thus, the denominators are divisors of 385, and their least common multiple is 385.

According to the problem, we have

$$
\frac{x}{\frac{385}{a}}+\frac{y}{\frac{385}{b}}+\frac{z}{\frac{385}{c}}=\frac{674}{385}
$$

where $a, b, c$ are divisors of 385 (i.e., $385 = 5 \cdot 7 \cdot 11$), and thus must be odd.

From (1),

$$
a x + b y + c z = 674
$$

Since the right-hand side is even, the left-hand side must also be even. Since $a, b, c$ are odd, either $x, y, z$ are all even, or one is even and two are odd. In both cases, the sum of the numerators of the sought fractions, $x + y + z$, is an even number. Since $x + y + z$ is equal to the sum of the digits of the denominators, the sum of the digits of the denominators of the sought fractions is even.

If the smallest denominator were 35 or larger, the sum of the numerators, and thus the sum of the digits of the denominators, would need to be greater than $\frac{674}{11} \approx 61$, otherwise the sum of the three fractions would be less than $\frac{674}{385}$. However, the sum of the digits of the denominators is at most 48 (if all three denominators are 385).

Therefore, the smallest denominator must be less than 35, i.e., using (a), it must be 11 or smaller.

Considering (a), (b), and (c), the three denominators could be:

| 1) | 5, | 7, | 11 | 5) | 11, | 35, | 35 | 9) | 11, | 55, | 77 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 2) | 5, | 7, | 55 | 6) | 11, | 35, | 55 | 10) | 11, | 55, | 385 |
| 3) | 5, | 7, | 77 | 7) | 11, | 35, | 77 | 11) | 11, | 77, | 385 |
| 4) | 5, | 7, | 385 | 8) | 11, | 35, | 385 | 12) | 11, | 385, | 385 |

These 12 possible cases need to be tried.

1) 

$$
\frac{x}{5} + \frac{y}{7} + \frac{z}{11} = \frac{674}{385}, \quad x + y + z = 14
$$

Subtracting 35 times the second equation from 385 times the first equation and dividing by 2, we get

$$
\begin{gathered}
21 x + 10 y = 92 \\
y = \frac{92 - 21 x}{10} = 9 - 2 x + \frac{2 - x}{10} = 9 - 2 x + t \\
x = 2 - 10 t
\end{gathered}
$$

Substituting back,

$$
\begin{gathered}
y = 9 - 4 + 20 t + t = 5 + 21 t \\
z = 14 - x - y = 14 - 2 + 10 t - 5 - 21 t = 7 - 11 t \\
x > 0, y > 0, z > 0, \quad \text{if } t = 0, \quad \text{then } \quad x = 2, y = 5, z = 7
\end{gathered}
$$

2) 

$$
\frac{x}{5} + \frac{y}{7} + \frac{z}{55} = \frac{674}{385}, \quad x + y + z = 22
$$

Eliminating $z$ and simplifying,

$$
35 x + 24 y = 260
$$

From this, the parametric solution for $x, y, z$ is:

$$
\begin{gathered}
x = 4 - 24 u, y = 5 + 35 u, z = 13 - 11 u \\
x > 0, y > 0, z > 0 \quad \text{if } u = 0; \quad \text{then } \quad x = 4, y = 5, z = 13
\end{gathered}
$$

3) 

$$
\frac{x}{5} + \frac{y}{7} + \frac{z}{77} = \frac{674}{385}, \quad x + y + z = 26
$$

Eliminating $z$ and simplifying,

$$
36 x + 25 y = 272
$$

From this, the parametric solution for $x, y, z$ is:

$$
\begin{gathered}
x = 2 - 25 u, \quad y = 8 + 36 u, \quad z = 16 - 11 u \\
x > 0, y > 0, z > 0, \quad \text{if } \quad u = 0: \quad \text{then } \quad x = 2, y = 8, z = 16
\end{gathered}
$$

In cases 4), 5), 6), 7), 10), 11), and 12), there are no positive integer solutions.

In case 8), $x = 17, y = 7, z = 2$, but the fraction $\frac{7}{35}$ is reducible, so it is not considered a solution.

Thus, case 9) remains.

9) 

$$
\frac{x}{11} + \frac{y}{55} + \frac{z}{77} = \frac{674}{385}, \quad x + y + z = 26
$$

Eliminating $z$ and simplifying,

$$
15 x + y = 272
$$

From this,

$$
\begin{gathered}
y = 272 - 15 x \\
z = 26 - x - 272 + 15 x = 14 x - 246 \\
x > 0, y > 0, z > 0, \quad \text{if } x = 18, \quad \text{then } \quad y = 2 \quad \text{and } z = 6
\end{gathered}
$$

Therefore, the problem has four solutions under the given constraints:

$$
\frac{2}{5} + \frac{5}{7} + \frac{7}{11} = \frac{674}{385}; \quad \frac{4}{5} + \frac{5}{7} + \frac{13}{55} = \frac{674}{385}; \quad \frac{2}{5} + \frac{8}{7} + \frac{16}{77} = \frac{674}{385}; \quad \frac{18}{11} + \frac{2}{55} + \frac{6}{77} = \frac{674}{385}
$$

Imre Csiszár (Bp., I., Petőfi g. II. o. t.)