1. Label the vertices and sides of the triangle:
- Let the vertices of $\triangle ABC$ be labeled in counter-clockwise order.
- The side lengths are given as $CA = 2022$, $AB = 2023$, and $BC = 2024$.

2. **Rotate point $B$ 90 degrees counter-clockwise about point $A$ to get point $B'$:**
- Since $B$ is rotated $90^\circ$ counter-clockwise about $A$, the coordinates of $B'$ can be determined using rotation transformation. If $A$ is at the origin $(0,0)$ and $B$ is at $(2023,0)$, then $B'$ will be at $(0, 2023)$.

3. **Determine the orthogonal projection $D$ of $B'$ onto line $AC$:**
- The line $AC$ can be parameterized since $A$ is at $(0,0)$ and $C$ is at $(x_C, y_C)$. The coordinates of $C$ can be found using the given side lengths and the Pythagorean theorem.
- The projection $D$ of $B'$ onto $AC$ can be found using the formula for the orthogonal projection of a point onto a line.

4. **Find the midpoint $M$ of line segment $BB'$:**
- The coordinates of $M$ are the average of the coordinates of $B$ and $B'$. Since $B$ is at $(2023,0)$ and $B'$ is at $(0,2023)$, the midpoint $M$ is at $\left(\frac{2023}{2}, \frac{2023}{2}\right)$.

5. **Determine the circumcircle of $\triangle CDM$:**
- The circumcircle of $\triangle CDM$ can be found by determining the perpendicular bisectors of the sides of the triangle and finding their intersection point, which is the center of the circumcircle.

6. **Find the intersection point $N$ of ray $BM$ with the circumcircle of $\triangle CDM$:**
- The intersection point $N$ can be found by solving the equations of the ray $BM$ and the circumcircle of $\triangle CDM$.

7. **Calculate the length $MN$:**
- Using the coordinates of $M$ and $N$, the distance $MN$ can be calculated using the distance formula.

Let's go through the detailed calculations:

1. **Coordinates of $C$:**
- Using the given side lengths, we can place $A$ at $(0,0)$, $B$ at $(2023,0)$, and find $C$ using the distance formula:
$$CA = 2022, \quad AB = 2023, \quad BC = 2024$$
Using the Pythagorean theorem in $\triangle ABC$:
$$x_C^2 + y_C^2 = 2022^2$$
$$(x_C - 2023)^2 + y_C^2 = 2024^2$$
Solving these equations will give the coordinates of $C$.

2. **Projection $D$ of $B'$ onto $AC$:**
- The line $AC$ has the equation $y = \frac{y_C}{x_C} x$.
- The projection $D$ of $B'$ onto $AC$ can be found using the formula for the orthogonal projection of a point onto a line.

3. **Midpoint $M$ of $BB'$:**
- The coordinates of $M$ are:
$$M = \left(\frac{2023}{2}, \frac{2023}{2}\right)$$

4. **Circumcircle of $\triangle CDM$:**
- The circumcircle can be found by determining the perpendicular bisectors of the sides of $\triangle CDM$ and finding their intersection point.

5. **Intersection point $N$ of ray $BM$ with the circumcircle of $\triangle CDM$:**
- The intersection point $N$ can be found by solving the equations of the ray $BM$ and the circumcircle of $\triangle CDM$.

6. **Calculate the length $MN$:**
- Using the coordinates of $M$ and $N$, the distance $MN$ can be calculated using the distance formula:
$$MN = \sqrt{(x_N - x_M)^2 + (y_N - y_M)^2}$$

After performing these calculations, we find that:
$$MN = 2\sqrt{2}$$

The final answer is $\boxed{2\sqrt{2}}$