An ending digit of $0$ requires a paired divisors of $2$ and $5$. Because $2$ appears far greater than $5$ does, therefore, it is sufficient to just count number of $5$ as divisor of $n$. It follows that this problem is equivalent to finding the four least $n$ which satisfies the following relation $$3\times\Big(\left\lfloor{\frac{n}{5}}\right\rfloor + \left\lfloor{\frac{n}{5^2}}\right\rfloor + \left\lfloor{\frac{n}{5^3}}\right\rfloor + \cdots\Big)=\left\lfloor{\frac{2n}{5}}\right\rfloor + \left\lfloor{\frac{2n}{5^2}}\right\rfloor + \left\lfloor{\frac{2n}{5^3}}\right\rfloor + \cdots$$

Solving such an equation typically requires some techniques (e.g. finding boundaries etc). However, in this case, it is not difficult to just try different values starting from $5$. The four least solutions are $8$, $9$, $13$, and $14$. Their sum is $44$. Therefore, the answer is $4+4=\boxed{8}$.