Find the number of ordered pairs $(a, b)$ of positive integers that are solutions of the following equation: \[a^2 + b^2 = ab(a+b).\]
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 \(\boxed{1}\)