Let $ \{a_n\}_{n=1}^\infty $ be an arithmetic progression  with $ a_1 > 0 $ and $ 5\cdot a_{13} = 6\cdot a_{19} $ . What is the smallest integer $ n$ such that $ a_n<0 $?
1. Let $\{a_n\}_{n=1}^\infty$ be an arithmetic progression with first term $a_1$ and common difference $d$. The general term of the arithmetic progression can be written as:
   \[
   a_n = a_1 + (n-1)d
   \]

2. Given that $a_1 > 0$ and $5 \cdot a_{13} = 6 \cdot a_{19}$, we can express $a_{13}$ and $a_{19}$ in terms of $a_1$ and $d$:
   \[
   a_{13} = a_1 + 12d
   \]
   \[
   a_{19} = a_1 + 18d
   \]

3. Substitute these expressions into the given equation:
   \[
   5(a_1 + 12d) = 6(a_1 + 18d)
   \]

4. Expand and simplify the equation:
   \[
   5a_1 + 60d = 6a_1 + 108d
   \]

5. Rearrange the terms to isolate $a_1$:
   \[
   5a_1 + 60d - 6a_1 - 108d = 0
   \]
   \[
   -a_1 - 48d = 0
   \]
   \[
   a_1 = -48d
   \]

6. Since $a_1 > 0$, it follows that $d < 0$ (because $a_1 = -48d$ and $a_1$ is positive).

7. We need to find the smallest integer $n$ such that $a_n < 0$. Using the general term formula:
   \[
   a_n = a_1 + (n-1)d
   \]

8. Substitute $a_1 = -48d$ into the equation:
   \[
   a_n = -48d + (n-1)d
   \]

9. Set $a_n < 0$ and solve for $n$:
   \[
   -48d + (n-1)d < 0
   \]
   \[
   -48d + nd - d < 0
   \]
   \[
   (n-49)d < 0
   \]

10. Since $d < 0$, divide both sides by $d$ (note that dividing by a negative number reverses the inequality):
    \[
    n - 49 > 0
    \]
    \[
    n > 49
    \]

11. The smallest integer $n$ that satisfies this inequality is $n = 50$.

The final answer is $\boxed{50}$.