For a positive number $n$,  we write $d (n)$ for the number of positive divisors of $n$. 
Determine all positive integers $k$ for which exist positive integers $a$ and $b$  with the property $k = d (a) = d (b) = d (2a + 3b)$.
1. **Given**: We need to determine all positive integers \( k \) for which there exist positive integers \( a \) and \( b \) such that \( k = d(a) = d(b) = d(2a + 3b) \), where \( d(n) \) denotes the number of positive divisors of \( n \).

2. **Step 1**: Consider the case where \( k \) is an even positive integer. Let \( k = 2l \) for some integer \( l \geq 0 \). We need to find \( a \) and \( b \) such that \( d(a) = d(b) = d(2a + 3b) = 2l \).

3. **Step 2**: Choose \( a = 2 \cdot 5^l \) and \( b = 3 \cdot 5^l \). Calculate the number of divisors for \( a \), \( b \), and \( 2a + 3b \):
   \[
   a = 2 \cdot 5^l \implies d(a) = d(2 \cdot 5^l) = (1+1)(l+1) = 2(l+1)
   \]
   \[
   b = 3 \cdot 5^l \implies d(b) = d(3 \cdot 5^l) = (1+1)(l+1) = 2(l+1)
   \]
   \[
   2a + 3b = 2(2 \cdot 5^l) + 3(3 \cdot 5^l) = 4 \cdot 5^l + 9 \cdot 5^l = 13 \cdot 5^l \implies d(2a + 3b) = d(13 \cdot 5^l) = (1+1)(l+1) = 2(l+1)
   \]
   Thus, \( d(a) = d(b) = d(2a + 3b) = 2(l+1) \), which is an even number.

4. **Step 3**: Now consider the case where \( k \) is an odd positive integer. Suppose \( k \) is odd, then \( k = 2m + 1 \) for some integer \( m \geq 0 \). If \( k \) is odd, then \( a \), \( b \), and \( 2a + 3b \) must all be perfect squares because the number of divisors of a perfect square is always odd.

5. **Step 4**: Assume \( a = m^2 \), \( b = n^2 \), and \( 2a + 3b = p^2 \). Then:
   \[
   2m^2 + 3n^2 = p^2
   \]
   Taking this equation modulo 3, we get:
   \[
   2m^2 \equiv p^2 \pmod{3}
   \]
   The quadratic residues modulo 3 are 0 and 1. Therefore, \( m \equiv 0 \pmod{3} \) and \( p \equiv 0 \pmod{3} \). Let \( m = 3m_1 \) and \( p = 3p_1 \), then:
   \[
   2(3m_1)^2 + 3n^2 = (3p_1)^2 \implies 18m_1^2 + 3n^2 = 9p_1^2 \implies 6m_1^2 + n^2 = 3p_1^2
   \]
   Taking this equation modulo 3, we get:
   \[
   n^2 \equiv 0 \pmod{3}
   \]
   Thus, \( n \equiv 0 \pmod{3} \). Let \( n = 3n_1 \), then:
   \[
   6m_1^2 + (3n_1)^2 = 3p_1^2 \implies 6m_1^2 + 9n_1^2 = 3p_1^2 \implies 2m_1^2 + 3n_1^2 = p_1^2
   \]
   Repeating this process leads to an infinite descent, implying \( m = n = p = 0 \), which is not allowed since \( a \), \( b \), and \( 2a + 3b \) must be positive integers.

6. **Conclusion**: Therefore, the only possible values for \( k \) are even positive integers.

The final answer is \(\boxed{\text{all even positive integers}}\).