The vertices of the convex quadrilateral $ABCD$ lie on the parabola $y=x^2$. It is known that $ABCD$ is cyclic and $AC$ is a diameter of its circumcircle. Let $M$ and $N$ be the midpoints of the diagonals of $AC$ and $BD$ respectively. Find the length of the projection of the segment $MN$ on the axis $Oy$. 
1. **Identify the coordinates of points \(A\), \(C\), \(M\), and \(N\):**
   - Given \(A = (a, a^2)\) and \(C = (c, c^2)\).
   - The midpoint \(M\) of \(AC\) is:
     \[
     M = \left( \frac{a+c}{2}, \frac{a^2 + c^2}{2} \right)
     \]

2. **Determine the coordinates of points \(B\) and \(D\):**
   - Since \(ABCD\) is cyclic and \(AC\) is a diameter of the circumcircle, the center of the circumcircle is the midpoint of \(AC\), which is \(M\).
   - Points \(B\) and \(D\) lie on the parabola \(y = x^2\) and the circle centered at \(M\) with radius \(\frac{|AC|}{2}\).
   - The equation of the circle is:
     \[
     \left( x - \frac{a+c}{2} \right)^2 + \left( y - \frac{a^2 + c^2}{2} \right)^2 = \left( \frac{|AC|}{2} \right)^2
     \]
   - The length of \(AC\) is:
     \[
     |AC| = \sqrt{(a - c)^2 + (a^2 - c^2)^2} = \sqrt{(a - c)^2 + (a - c)(a + c)^2} = \sqrt{(a - c)^2 (1 + (a + c)^2)}
     \]
   - Therefore, the radius squared is:
     \[
     \left( \frac{|AC|}{2} \right)^2 = \frac{(a - c)^2 (1 + (a + c)^2)}{4}
     \]

3. **Solve for \(B\) and \(D\):**
   - The coordinates of \(B\) and \(D\) must satisfy both the parabola equation \(y = x^2\) and the circle equation.
   - Substituting \(y = x^2\) into the circle equation:
     \[
     \left( x - \frac{a+c}{2} \right)^2 + \left( x^2 - \frac{a^2 + c^2}{2} \right)^2 = \frac{(a - c)^2 (1 + (a + c)^2)}{4}
     \]
   - Simplifying, we get:
     \[
     (x - \frac{a+c}{2})^2 + (x^2 - \frac{a^2 + c^2}{2})^2 = \frac{(a - c)^2 (1 + (a + c)^2)}{4}
     \]
   - This simplifies to:
     \[
     (x - a)(x - c)(x^2 + (a+c)x + ac + 1) = 0
     \]
   - Therefore, the roots \(b\) and \(d\) of the quadratic equation are:
     \[
     x^2 + (a+c)x + ac + 1 = 0
     \]

4. **Calculate the coordinates of \(N\):**
   - The roots \(b\) and \(d\) satisfy:
     \[
     b + d = -(a + c) \quad \text{and} \quad bd = ac + 1
     \]
   - The midpoint \(N\) of \(BD\) is:
     \[
     N = \left( \frac{b+d}{2}, \frac{b^2 + d^2}{2} \right)
     \]
   - Using the identity \(b^2 + d^2 = (b+d)^2 - 2bd\):
     \[
     b^2 + d^2 = (-(a+c))^2 - 2(ac + 1) = a^2 + c^2 + 2ac - 2ac - 2 = a^2 + c^2 - 2
     \]
   - Therefore, \(N\) is:
     \[
     N = \left( \frac{-(a+c)}{2}, \frac{a^2 + c^2 - 2}{2} \right)
     \]

5. **Calculate the projection of \(MN\) on the \(y\)-axis:**
   - The \(y\)-coordinates of \(M\) and \(N\) are \(\frac{a^2 + c^2}{2}\) and \(\frac{a^2 + c^2 - 2}{2}\) respectively.
   - The projection of \(MN\) on the \(y\)-axis is:
     \[
     \left| \frac{a^2 + c^2 - 2}{2} - \frac{a^2 + c^2}{2} \right| = \left| \frac{a^2 + c^2 - 2 - a^2 - c^2}{2} \right| = \left| \frac{-2}{2} \right| = 1
     \]

The final answer is \(\boxed{1}\).