Solution:

The answer is $(\mathbf{D})$. We have $18000=2^{4} \cdot 3^{2} \cdot 5^{3}$, so all divisors of $18000$ are of the form $2^{a} \cdot 3^{b} \cdot 5^{c}$, where $a \in\{0,1,2,3,4\}$, $b \in\{0,1,2\}$ and $c \in\{0,1,2,3\}$. In order for the decimal representation of the divisor to end in $50$, the number must be divisible by $50$, hence $a \geq 1$ and $c \geq 2$. Note, moreover, that if $a \geq 2$ and $c \geq 2$, then the decimal representation of the divisor ends in the digits "00". It is therefore also necessary to require $a=1$. Hence the required sum is given by
$$\sum_{b=0}^{2} \sum_{c=2}^{3} 2 \cdot 3^{b} \cdot 5^{c} = 2 \cdot \left(\sum_{b=0}^{2} 3^{b}\right) \cdot \left(\sum_{c=2}^{3} 5^{c}\right) = 2 \cdot 13 \cdot 150 = 3900.$$