For which values of the real parameter $r$ the equation $r^2 x^2+2rx+4=28r^2$ has two distinct integer roots?
1. **Rewrite the given equation:**
   \[
   r^2 x^2 + 2rx + 4 = 28r^2
   \]
   Subtract \(28r^2\) from both sides to form a standard quadratic equation:
   \[
   r^2 x^2 + 2rx + 4 - 28r^2 = 0
   \]
   Simplify:
   \[
   r^2 x^2 + 2rx + 4 - 28r^2 = r^2 x^2 + 2rx - 24r^2 + 4 = 0
   \]
   \[
   r^2 x^2 + 2rx + 4(1 - 7r^2) = 0
   \]

2. **Identify the quadratic equation in \(x\):**
   \[
   r^2 x^2 + 2rx + 4(1 - 7r^2) = 0
   \]
   This is a quadratic equation of the form \(ax^2 + bx + c = 0\) where:
   \[
   a = r^2, \quad b = 2r, \quad c = 4(1 - 7r^2)
   \]

3. **Conditions for two distinct integer roots:**
   - The discriminant \(\Delta\) of the quadratic equation must be a perfect square.
   - The discriminant \(\Delta\) is given by:
     \[
     \Delta = b^2 - 4ac
     \]
     Substitute \(a\), \(b\), and \(c\):
     \[
     \Delta = (2r)^2 - 4(r^2)(4(1 - 7r^2)) = 4r^2 - 16r^2(1 - 7r^2)
     \]
     Simplify:
     \[
     \Delta = 4r^2 - 16r^2 + 112r^4 = -12r^2 + 112r^4
     \]
     Factor out \(4r^2\):
     \[
     \Delta = 4r^2(28r^2 - 3)
     \]
     For \(\Delta\) to be a perfect square:
     \[
     4r^2(28r^2 - 3) = k^2 \quad \text{for some integer } k
     \]

4. **Sum and product of roots:**
   - The sum of the roots \(S\) is given by:
     \[
     S = -\frac{b}{a} = -\frac{2r}{r^2} = -\frac{2}{r}
     \]
   - The product of the roots \(P\) is given by:
     \[
     P = \frac{c}{a} = \frac{4(1 - 7r^2)}{r^2} = \frac{4}{r^2} - 28
     \]

5. **Determine possible values of \(r\):**
   - Since the roots are integers, let the roots be \(p\) and \(q\). Then:
     \[
     p + q = -\frac{2}{r} \quad \text{and} \quad pq = \frac{4}{r^2} - 28
     \]
   - For \(p + q\) to be an integer, \(-\frac{2}{r}\) must be an integer. Let \(-\frac{2}{r} = n\), where \(n\) is an integer. Then:
     \[
     r = -\frac{2}{n}
     \]
   - Substitute \(r = -\frac{2}{n}\) into the product of the roots:
     \[
     pq = \frac{4}{\left(-\frac{2}{n}\right)^2} - 28 = \frac{4n^2}{4} - 28 = n^2 - 28
     \]
   - For \(pq\) to be an integer, \(n^2 - 28\) must be an integer. Therefore, \(n^2\) must be greater than 28.

6. **Find integer values of \(n\):**
   - \(n^2 < \frac{112}{3} \approx 37.33\)
   - Therefore, \(1 \leq |n| \leq 6\)

7. **Check each value of \(n\):**
   - For \(n = \pm 1\):
     \[
     x^2 \pm x - 27 = 0 \quad \text{(no integer roots)}
     \]
   - For \(n = \pm 2\):
     \[
     x^2 \pm 2x - 24 = 0 \quad \text{(integer roots: 6 and -4)}
     \]
   - For \(n = \pm 3\):
     \[
     x^2 \pm 3x - 19 = 0 \quad \text{(no integer roots)}
     \]
   - For \(n = \pm 4\):
     \[
     x^2 \pm 4x - 12 = 0 \quad \text{(integer roots: 6 and -2)}
     \]
   - For \(n = \pm 5\):
     \[
     x^2 \pm 5x - 3 = 0 \quad \text{(no integer roots)}
     \]
   - For \(n = \pm 6\):
     \[
     x^2 \pm 6x + 8 = 0 \quad \text{(integer roots: -2 and -4)}
     \]

8. **Determine corresponding values of \(r\):**
   - For \(n = \pm 2\), \(r = \pm \frac{1}{2}\)
   - For \(n = \pm 4\), \(r = \pm \frac{1}{4}\)
   - For \(n = \pm 6\), \(r = \pm \frac{1}{6}\)

The final answer is \(\boxed{r \in \left\{ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3} \right\}}\)