Let $s(n)$ be the number of positive integer divisors of $n$. 
Find the all positive values of $k$ that is providing $k=s(a)=s(b)=s(2a+3b)$.
1. We start by noting that \( s(n) \) represents the number of positive integer divisors of \( n \). We are given that \( k = s(a) = s(b) = s(2a + 3b) \).

2. If \( k \) is odd, then \( a \) and \( b \) must be perfect squares. This is because the number of divisors of a perfect square is always odd. Let \( a = x^2 \) and \( b = y^2 \).

3. We need \( 2a + 3b \) to also be a perfect square. Let \( 2a + 3b = z^2 \). Substituting \( a = x^2 \) and \( b = y^2 \), we get:
   \[
   2x^2 + 3y^2 = z^2
   \]

4. Consider the equation modulo 3:
   \[
   2x^2 + 3y^2 \equiv z^2 \pmod{3}
   \]
   Since \( 3y^2 \equiv 0 \pmod{3} \), we have:
   \[
   2x^2 \equiv z^2 \pmod{3}
   \]

5. The quadratic residues modulo 3 are 0 and 1. Therefore, \( x^2 \equiv 0 \) or \( 1 \pmod{3} \). If \( x^2 \equiv 0 \pmod{3} \), then \( x \equiv 0 \pmod{3} \). Similarly, \( z^2 \equiv 0 \pmod{3} \) implies \( z \equiv 0 \pmod{3} \).

6. Let \( x = 3k \) and \( z = 3m \). Substituting these into the equation, we get:
   \[
   2(3k)^2 + 3y^2 = (3m)^2
   \]
   Simplifying, we obtain:
   \[
   18k^2 + 3y^2 = 9m^2
   \]
   Dividing through by 3, we get:
   \[
   6k^2 + y^2 = 3m^2
   \]

7. Consider the equation modulo 3 again:
   \[
   y^2 \equiv 0 \pmod{3}
   \]
   This implies \( y \equiv 0 \pmod{3} \). Let \( y = 3n \). Substituting this into the equation, we get:
   \[
   6k^2 + 9n^2 = 3m^2
   \]
   Dividing through by 3, we get:
   \[
   2k^2 + 3n^2 = m^2
   \]

8. The equation \( 2k^2 + 3n^2 = m^2 \) is a form of a Diophantine equation. By Fermat's method of infinite descent, there are no non-trivial integer solutions to this equation. Therefore, there are no solutions for \( k \) being odd.

9. Since \( k \) cannot be odd, we conclude that \( k \) must be even.

The final answer is \( \boxed{ k } \) must be even.