Let $ABC$ be an equilateral triangle with side 3. A circle $C_1$ is tangent to $AB$ and $AC$.
A circle $C_2$, with a radius smaller than the radius of $C_1$, is tangent to $AB$ and $AC$ as well as externally tangent to $C_1$.
Successively, for $n$ positive integer, the circle $C_{n+1}$, with a radius smaller than the radius of $C_n$, is tangent to $AB$ and $AC$ and is externally tangent to $C_n$.
Determine the possible values for the radius of $C_1$ such that 4 circles from this sequence, but not 5, are contained on the interior of the triangle $ABC$.
1. **Identify the position of the circle centers:**
   The centers of the circles \(C_1, C_2, \ldots\) must lie on the internal bisector \(AM\) of \(\angle BAC\), since their distances from the sides \(AB\) and \(AC\) are equal. 

2. **Establish the relationship between the radii:**
   Consider the first two circles \(C_1\) and \(C_2\). Let the radius of \(C_1\) be \(x\) and the radius of \(C_2\) be \(y\). Since \(C_2\) is externally tangent to \(C_1\) and both are tangent to \(AB\) and \(AC\), the centers of these circles form a right triangle with the internal bisector \(AM\). The hypotenuse of this triangle is \(x + y\) and the leg opposite the \(30^\circ\) angle is \(y - x\). Using the ratio for a \(30^\circ-60^\circ-90^\circ\) triangle, we have:
   \[
   \frac{y - x}{x + y} = \frac{1}{2}
   \]
   Solving for \(y\), we get:
   \[
   2(y - x) = x + y \implies 2y - 2x = x + y \implies 2y - y = x + 2x \implies y = 3x
   \]

3. **Iterate the relationship for subsequent circles:**
   For \(n = 2, 3, \ldots\), the radius of the \(n\)-th circle \(C_n\) can be expressed as \(r_n = 3^{n-1}x\). The distance from \(A\) to the center of the \(n\)-th circle \(O_n\) is the sum of the radii of all previous circles plus the radius of the \(n\)-th circle:
   \[
   \overline{AP_n} = x + (x + 3x) + (3x + 9x) + \cdots + (3^{n-2}x + 3^{n-1}x) + 3^{n-1}x
   \]
   Simplifying, we get:
   \[
   \overline{AP_n} = x \left(1 + 4 + 12 + \cdots + 3^{n-1} + 3^{n-1}\right)
   \]
   This is a geometric series with the first term \(a = 1\) and common ratio \(r = 3\):
   \[
   \overline{AP_n} = x \left(1 + 4 + 12 + \cdots + 3^{n-1} + 3^{n-1}\right) = x \left(1 + 3 \sum_{k=0}^{n-2} 3^k\right)
   \]
   The sum of the geometric series is:
   \[
   \sum_{k=0}^{n-2} 3^k = \frac{3^{n-1} - 1}{2}
   \]
   Therefore:
   \[
   \overline{AP_n} = x \left(1 + 3 \cdot \frac{3^{n-1} - 1}{2}\right) = x \left(1 + \frac{3^{n} - 3}{2}\right) = x \left(\frac{3^n + 1}{2}\right)
   \]

4. **Determine the condition for 4 circles to fit inside the triangle:**
   The length of the internal bisector \(AM\) of \(\angle BAC\) in an equilateral triangle with side length 3 is:
   \[
   AM = \frac{\sqrt{3}}{2} \cdot 3 = \frac{3\sqrt{3}}{2}
   \]
   We need:
   \[
   \overline{AP_4} < \overline{AM} < \overline{AP_5}
   \]
   Substituting the expressions for \(\overline{AP_4}\) and \(\overline{AP_5}\):
   \[
   x \left(\frac{3^4 + 1}{2}\right) < \frac{3\sqrt{3}}{2} < x \left(\frac{3^5 + 1}{2}\right)
   \]
   Simplifying:
   \[
   x \left(\frac{81 + 1}{2}\right) < \frac{3\sqrt{3}}{2} < x \left(\frac{243 + 1}{2}\right)
   \]
   \[
   x \cdot 41 < \frac{3\sqrt{3}}{2} < x \cdot 122
   \]
   Solving for \(x\):
   \[
   x < \frac{3\sqrt{3}}{82} \quad \text{and} \quad x > \frac{3\sqrt{3}}{244}
   \]
   Therefore, the possible values for \(x\) are:
   \[
   \frac{3\sqrt{3}}{244} < x < \frac{3\sqrt{3}}{82}
   \]

The final answer is \(\boxed{\frac{3\sqrt{3}}{244} < x < \frac{3\sqrt{3}}{82}}\).