To find the number of positive integers $x$ satisfying the given conditions, we need to analyze the divisibility condition $x^2 - x$ by $10^{2006}$.

1. Condition Analysis:
$$x^2 - x \equiv 0 \pmod{10^{2006}}$$
This implies:
$$x(x-1) \equiv 0 \pmod{10^{2006}}$$
Since $10^{2006} = 2^{2006} \times 5^{2006}$, we need:
$$x(x-1) \equiv 0 \pmod{2^{2006}} \quad \text{and} \quad x(x-1) \equiv 0 \pmod{5^{2006}}$$

2. **Divisibility by $2^{2006}$**:
$$x(x-1) \equiv 0 \pmod{2^{2006}}$$
This means either $x \equiv 0 \pmod{2^{2006}}$ or $x-1 \equiv 0 \pmod{2^{2006}}$. Therefore:
$$x \equiv 0 \pmod{2^{2006}} \quad \text{or} \quad x \equiv 1 \pmod{2^{2006}}$$

3. **Divisibility by $5^{2006}$**:
$$x(x-1) \equiv 0 \pmod{5^{2006}}$$
This means either $x \equiv 0 \pmod{5^{2006}}$ or $x-1 \equiv 0 \pmod{5^{2006}}$. Therefore:
$$x \equiv 0 \pmod{5^{2006}} \quad \text{or} \quad x \equiv 1 \pmod{5^{2006}}$$

4. Combining Conditions Using the Chinese Remainder Theorem:
We need to find the number of solutions $x$ that satisfy both sets of congruences:
- $x \equiv 0 \pmod{2^{2006}}$ and $x \equiv 0 \pmod{5^{2006}}$
- $x \equiv 0 \pmod{2^{2006}}$ and $x \equiv 1 \pmod{5^{2006}}$
- $x \equiv 1 \pmod{2^{2006}}$ and $x \equiv 0 \pmod{5^{2006}}$

Using the Chinese Remainder Theorem, we can solve these systems of congruences:
- For $x \equiv 0 \pmod{2^{2006}}$ and $x \equiv 0 \pmod{5^{2006}}$:
$$x \equiv 0 \pmod{10^{2006}}$$
This solution is not valid since $x$ must be positive and less than $10^{2006}$.

- For $x \equiv 0 \pmod{2^{2006}}$ and $x \equiv 1 \pmod{5^{2006}}$:
There exists exactly one solution $0 < x < 10^{2006}$.

- For $x \equiv 1 \pmod{2^{2006}}$ and $x \equiv 0 \pmod{5^{2006}}$:
There exists exactly one solution $0 < x < 10^{2006}$.

5. Counting the Solutions:
Including $x = 1$, we have:
- One solution from $x \equiv 0 \pmod{2^{2006}}$ and $x \equiv 1 \pmod{5^{2006}}$
- One solution from $x \equiv 1 \pmod{2^{2006}}$ and $x \equiv 0 \pmod{5^{2006}}$
- The solution $x = 1$

Therefore, there are a total of 3 solutions.

The final answer is $\boxed{3}$.