We first see that such an $N$ must be a multiple of $2$ since otherwise the units digits of the divisors would never be even. We also see that $N$ must be a multiple of $5$ since any number which has a units digit of $5$ must be a multiple of $5$. Thus $N$ is a multiple of $10$ and so we may write $$N = 2 \cdot 5k$$ for some positive integer $k$. Now we take two cases:
Case $1$: $3 \mid N$
In this case we prove that the minimal $N = 2 \cdot 3^3 \cdot 5$. If the exponent of $3$ is reduced, then the new prime $7$ must be added to the prime factorization of $N$ and so $3^2 \nmid N$ otherwise $N$ would be larger. But then we have the minimal such $N$ to be $2 \cdot 3 \cdot 5 \cdot 7$, which doesn't have a divisor with a units digit of $9$. Thus the exponent of $3$ cannot be reduced and so $N$ is minimal in this case.
Case $2$: $3 \nmid N$
In this case we prove that $N > 2 \cdot 3^3 \cdot 5$. We must add a new prime greater than or equal to $7$ into the prime factorization of $2 \cdot 5$. We also must add at least one other prime since otherwise we will have exactly $8$ divisors and it can't be $2$ since we can't get a units digit of $8$. Thus we have $N \geq 2 \cdot 5^2 \cdot 7$, which means that $N > 2 \cdot 3^3 \cdot 5$ since $7 \cdot 5 > 3^3$.