The perimeter of a triangle is a natural number, its circumradius is equal to $\frac{65}{8}$, and the inradius is equal to $4$. Find the sides of the triangle.
Given:
- The perimeter of the triangle \( P \) is a natural number.
- The circumradius \( R = \frac{65}{8} \).
- The inradius \( r = 4 \).

We need to find the sides \( a, b, c \) of the triangle.

1. **Using the formula for the area of the triangle:**
   \[
   \text{Area} = \Delta = r \cdot s
   \]
   where \( s \) is the semi-perimeter of the triangle:
   \[
   s = \frac{a + b + c}{2}
   \]
   Therefore:
   \[
   \Delta = 4 \cdot s
   \]

2. **Using the formula for the area of the triangle with circumradius:**
   \[
   \Delta = \frac{abc}{4R}
   \]
   Given \( R = \frac{65}{8} \), we have:
   \[
   \Delta = \frac{abc}{4 \cdot \frac{65}{8}} = \frac{abc \cdot 8}{260} = \frac{4abc}{130} = \frac{2abc}{65}
   \]

3. **Equating the two expressions for the area:**
   \[
   4s = \frac{2abc}{65}
   \]
   \[
   4 \left( \frac{a + b + c}{2} \right) = \frac{2abc}{65}
   \]
   \[
   2(a + b + c) = \frac{2abc}{65}
   \]
   \[
   65(a + b + c) = abc
   \]

4. **Using the given condition \( a, b, c < 2R < 17 \):**
   \[
   a, b, c < 17
   \]

5. **Assuming \( a = 13 \) (since \( 13 \) is a factor of \( 65 \)):**
   \[
   65(13 + b + c) = 13bc
   \]
   \[
   65 \cdot 13 + 65b + 65c = 13bc
   \]
   \[
   845 + 65b + 65c = 13bc
   \]
   \[
   65b + 65c - 13bc = -845
   \]
   \[
   13bc - 65b - 65c = 845
   \]
   \[
   bc - 5b - 5c = 65
   \]
   \[
   (b - 5)(c - 5) = 90
   \]

6. **Finding integer solutions for \( (b - 5)(c - 5) = 90 \):**
   - Possible pairs \((b-5, c-5)\) are \((1, 90), (2, 45), (3, 30), (5, 18), (6, 15), (9, 10)\).

7. **Checking valid pairs:**
   - For \( (b-5, c-5) = (6, 15) \):
     \[
     b = 11, \quad c = 20 \quad \text{(not valid since } c > 17\text{)}
     \]
   - For \( (b-5, c-5) = (9, 10) \):
     \[
     b = 14, \quad c = 15
     \]

8. **Verifying the solution:**
   - \( a = 13, b = 14, c = 15 \)
   - Perimeter \( P = a + b + c = 13 + 14 + 15 = 42 \)
   - Semi-perimeter \( s = \frac{42}{2} = 21 \)
   - Area \( \Delta = 4 \cdot 21 = 84 \)
   - Using Heron's formula:
     \[
     \Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056} = 84
     \]
   - Circumradius check:
     \[
     R = \frac{abc}{4\Delta} = \frac{13 \cdot 14 \cdot 15}{4 \cdot 84} = \frac{2730}{336} = \frac{65}{8}
     \]

Thus, the sides of the triangle are \( (13, 14, 15) \).

The final answer is \( \boxed{ (13, 14, 15) } \)