The diameter of a circle of radius $R$ is divided into $4$ equal parts. The point $M$ is taken on the circle. Prove that the sum of the squares of the distances from the point $M$ to the points of division (together with the ends of the diameter) does not depend on the choice of the point $M$. Calculate this sum.
1. **Set up the coordinate system:**
   Consider a coordinate system with the center of the circle at the origin \((0,0)\) and the diameter along the x-axis. The radius of the circle is \(R\).

2. **Identify the division points:**
   The diameter is divided into 4 equal parts, so the points of division (including the ends of the diameter) are:
   \[
   A = (-R, 0), \quad B = \left(-\frac{R}{2}, 0\right), \quad O = (0, 0), \quad C = \left(\frac{R}{2}, 0\right), \quad D = (R, 0)
   \]

3. **Express the coordinates of point \(M\):**
   Let \(M\) be any point on the circle. The coordinates of \(M\) can be written as:
   \[
   M = (R \cos(\theta), R \sin(\theta))
   \]
   where \(\theta\) is the angle that the radius to \(M\) makes with the positive x-axis.

4. **Calculate the squared distances from \(M\) to each division point:**
   - Distance from \(M\) to \(A\):
     \[
     MA^2 = \left(R \cos(\theta) + R\right)^2 + \left(R \sin(\theta) - 0\right)^2 = (R + R \cos(\theta))^2 + (R \sin(\theta))^2
     \]
     Simplifying:
     \[
     MA^2 = R^2 (1 + \cos(\theta))^2 + R^2 \sin^2(\theta)
     \]
     Using the Pythagorean identity \(\sin^2(\theta) + \cos^2(\theta) = 1\):
     \[
     MA^2 = R^2 (1 + 2\cos(\theta) + \cos^2(\theta)) + R^2 \sin^2(\theta) = R^2 (2 + 2\cos(\theta))
     \]
     \[
     MA^2 = 2R^2 (1 + \cos(\theta))
     \]

   - Distance from \(M\) to \(B\):
     \[
     MB^2 = \left(R \cos(\theta) + \frac{R}{2}\right)^2 + \left(R \sin(\theta) - 0\right)^2 = \left(\frac{R}{2} + R \cos(\theta)\right)^2 + (R \sin(\theta))^2
     \]
     Simplifying:
     \[
     MB^2 = R^2 \left(\frac{1}{4} + \cos(\theta) + \cos^2(\theta)\right) + R^2 \sin^2(\theta)
     \]
     Using the Pythagorean identity:
     \[
     MB^2 = R^2 \left(\frac{1}{4} + \cos(\theta) + 1\right) = R^2 \left(\frac{5}{4} + \cos(\theta)\right)
     \]

   - Distance from \(M\) to \(O\):
     \[
     MO^2 = (R \cos(\theta) - 0)^2 + (R \sin(\theta) - 0)^2 = R^2 \cos^2(\theta) + R^2 \sin^2(\theta)
     \]
     Using the Pythagorean identity:
     \[
     MO^2 = R^2
     \]

   - Distance from \(M\) to \(C\):
     \[
     MC^2 = \left(R \cos(\theta) - \frac{R}{2}\right)^2 + \left(R \sin(\theta) - 0\right)^2 = \left(\frac{R}{2} - R \cos(\theta)\right)^2 + (R \sin(\theta))^2
     \]
     Simplifying:
     \[
     MC^2 = R^2 \left(\frac{1}{4} - \cos(\theta) + \cos^2(\theta)\right) + R^2 \sin^2(\theta)
     \]
     Using the Pythagorean identity:
     \[
     MC^2 = R^2 \left(\frac{1}{4} - \cos(\theta) + 1\right) = R^2 \left(\frac{5}{4} - \cos(\theta)\right)
     \]

   - Distance from \(M\) to \(D\):
     \[
     MD^2 = \left(R \cos(\theta) - R\right)^2 + \left(R \sin(\theta) - 0\right)^2 = (R - R \cos(\theta))^2 + (R \sin(\theta))^2
     \]
     Simplifying:
     \[
     MD^2 = R^2 (1 - \cos(\theta))^2 + R^2 \sin^2(\theta)
     \]
     Using the Pythagorean identity:
     \[
     MD^2 = R^2 (1 - 2\cos(\theta) + \cos^2(\theta)) + R^2 \sin^2(\theta) = R^2 (2 - 2\cos(\theta))
     \]
     \[
     MD^2 = 2R^2 (1 - \cos(\theta))
     \]

5. **Sum the squared distances:**
   \[
   \sum_{i=A,B,O,C,D} MP_i^2 = MA^2 + MB^2 + MO^2 + MC^2 + MD^2
   \]
   Substituting the values:
   \[
   \sum_{i=A,B,O,C,D} MP_i^2 = 2R^2 (1 + \cos(\theta)) + R^2 \left(\frac{5}{4} + \cos(\theta)\right) + R^2 + R^2 \left(\frac{5}{4} - \cos(\theta)\right) + 2R^2 (1 - \cos(\theta))
   \]
   Simplifying:
   \[
   \sum_{i=A,B,O,C,D} MP_i^2 = 2R^2 + 2R^2 \cos(\theta) + \frac{5R^2}{4} + R^2 \cos(\theta) + R^2 + \frac{5R^2}{4} - R^2 \cos(\theta) + 2R^2 - 2R^2 \cos(\theta)
   \]
   \[
   \sum_{i=A,B,O,C,D} MP_i^2 = 2R^2 + \frac{5R^2}{4} + R^2 + \frac{5R^2}{4} + 2R^2
   \]
   \[
   \sum_{i=A,B,O,C,D} MP_i^2 = 2R^2 + \frac{10R^2}{4} + R^2 + 2R^2 = 2R^2 + 2.5R^2 + R^2 + 2R^2 = 7.5R^2
   \]
   \[
   \sum_{i=A,B,O,C,D} MP_i^2 = \frac{15R^2}{2}
   \]

The final answer is \(\boxed{\frac{15R^2}{2}}\)