To find the number of ordered pairs $(a, b)$ of positive integers that satisfy the equation $a^2 + b^2 = ab(a + b)$, we will follow these steps:

1. Rearrange the given equation:
$$a^2 + b^2 = ab(a + b)$$
We can rewrite this equation as:
$$a^2 + b^2 = a^2b + ab^2$$

2. Move all terms to one side of the equation:
$$a^2 + b^2 - a^2b - ab^2 = 0$$

3. Factor the equation:
Notice that we can factor by grouping:
$$a^2 + b^2 - a^2b - ab^2 = a^2(1 - b) + b^2(1 - a) = 0$$

4. Analyze the factored form:
For the equation $a^2(1 - b) + b^2(1 - a) = 0$ to hold, we need to consider the cases where each term is zero.

- Case 1: $a^2(1 - b) = 0$
$$a^2 = 0 \quad \text{or} \quad 1 - b = 0$$
Since $a$ is a positive integer, $a^2 \neq 0$. Therefore, $1 - b = 0$ which implies $b = 1$.

- Case 2: $b^2(1 - a) = 0$
$$b^2 = 0 \quad \text{or} \quad 1 - a = 0$$
Since $b$ is a positive integer, $b^2 \neq 0$. Therefore, $1 - a = 0$ which implies $a = 1$.

5. Check the solution:
The only solution from the above cases is $a = 1$ and $b = 1$. We can verify this by substituting $a = 1$ and $b = 1$ back into the original equation:
$$1^2 + 1^2 = 1 \cdot 1 \cdot (1 + 1)$$
$$2 = 2$$
This confirms that $(1, 1)$ is indeed a solution.

6. Consider other possible values:
Since $a$ and $b$ are positive integers, and we have shown that for $a > 1$ or $b > 1$, the equation does not hold, there are no other solutions.

Conclusion:
The only ordered pair $(a, b)$ that satisfies the equation is $(1, 1)$.

The final answer is $\boxed{1}$