What is the least positive integer that is divisorous?

$2\cdot 3^3\cdot 5=270$.

From 1,3,5,$3^3=27$, and $3^2=9$, we get all the odd digits, and multiplying by two, we get all the even ones. We can see that all the prime factors here are necessary, and by minimality of the prime factors in question, we cannot get a smaller one.

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 $7 \cdot 5 > 3^3$.