Calculate the sum of the ratio parts.

The ratio of the sides of the triangle is $2:3:4$.
The sum of the ratio parts is:

$$2 + 3 + 4 = 9$$

Let the common multiplier for the ratio be $x$. The perimeter of the triangle is $120 \text{ mm}$.
The sum of the sides is equal to the perimeter:

$$2x + 3x + 4x = 120 \text{ mm}$$ $$9x = 120 \text{ mm}$$ $$x = \frac{120}{9} \text{ mm} = \frac{40}{3} \text{ mm}$$

Now, calculate the length of each side:
Side 1: $2x = 2 \times \frac{40}{3} \text{ mm} = \frac{80}{3} \text{ mm} \approx 26.67 \text{ mm}$
Side 2: $3x = 3 \times \frac{40}{3} \text{ mm} = 40 \text{ mm}$
Side 3: $4x = 4 \times \frac{40}{3} \text{ mm} = \frac{160}{3} \text{ mm} \approx 53.33 \text{ mm}$

To construct the triangle with sides $\frac{80}{3} \text{ mm}$, $40 \text{ mm}$, and $\frac{160}{3} \text{ mm}$:

1. Draw a line segment, say $AB$, with a length of $40 \text{ mm}$. This will be the base of your triangle.
2. Place the compass point at $A$ and open it to a radius of $\frac{80}{3} \text{ mm}$ (approximately $26.67 \text{ mm}$). Draw an arc above the line segment $AB$.
3. Place the compass point at $B$ and open it to a radius of $\frac{160}{3} \text{ mm}$ (approximately $53.33 \text{ mm}$). Draw another arc above the line segment $AB$, intersecting the first arc.
4. Label the point where the two arcs intersect as $C$.
5. Draw line segments $AC$ and $BC$. Triangle $ABC$ is the required triangle.

The lengths of the sides of the triangle are $\boxed{\text{Side 1} = \frac{80}{3} \text{ mm}, \text{Side 2} = 40 \text{ mm}, \text{Side 3} = \frac{160}{3} \text{ mm}}$.