Side $AB$ of triangle $ABC$ was divided into $n$ equal parts (dividing points $B_0 = A, B_1, B_2, ..., B_n = B$), and side $AC$ of this triangle was divided into $(n + 1)$ equal parts (dividing points $C_0 = A, C_1, C_2, ..., C_{n+1} = C$). Colored are the triangles $C_iB_iC_{i+1}$ (where $i = 1,2, ..., n$). What part of the area of the triangle is painted over?
1. **Base Case:**
   For \( n = 1 \), the triangle \( \triangle C_1B_1C_2 \) is formed by dividing \( AB \) into 1 part and \( AC \) into 2 parts. It is easy to see that the area of \( \triangle C_1B_1C_2 \) is half of the area of \( \triangle ABC \).

2. **Induction Hypothesis:**
   Assume that for \( n = k \), the shaded area is \( \frac{1}{2} \) of the area of \( \triangle ABC \).

3. **Induction Step:**
   For \( n = k+1 \), we need to show that the shaded area is still \( \frac{1}{2} \) of the area of \( \triangle ABC \).

   - The triangles \( \triangle C_1B_1C_2, \triangle C_2B_2C_3, \ldots, \triangle C_kB_kC_{k+1} \) are already shaded and their total area is \( \frac{1}{2} \) of the area of \( \triangle ABC \) by the induction hypothesis.
   - We need to add the area of the new triangle \( \triangle C_{k+1}B_{k+1}C_{k+2} \).

4. **Area Calculation:**
   - The area of \( \triangle C_{k+1}B_{k+1}C_{k+2} \) can be calculated as follows:
     \[
     \text{Area of } \triangle C_{k+1}B_{k+1}C_{k+2} = \frac{1}{k+2} \times \text{Area of } \triangle ABC
     \]
   - The total shaded area for \( n = k+1 \) is:
     \[
     \left( \frac{1}{2} \times \text{Area of } \triangle ABC \right) + \left( \frac{1}{k+2} \times \text{Area of } \triangle ABC \right)
     \]
   - Simplifying the expression:
     \[
     \frac{1}{2} \times \text{Area of } \triangle ABC + \frac{1}{k+2} \times \text{Area of } \triangle ABC = \frac{k+2}{2(k+2)} \times \text{Area of } \triangle ABC = \frac{1}{2} \times \text{Area of } \triangle ABC
     \]

5. **Conclusion:**
   By induction, the shaded area is always \( \frac{1}{2} \) of the area of \( \triangle ABC \) for any \( n \).

\(\blacksquare\)

The final answer is \( \boxed{ \frac{1}{2} } \) of the area of \(\triangle ABC\).