Identify the digits and their positions

The digits are: $1,1,2,2,2,2,3,4,4$

Total digits $=9$, so each number formed has $9$ places.

Odd digits are: $1,1,3$ There are $3$ odd digits.

Even places among $9$ positions are: $2,4,6,8$ So there are $4$ even positions.

Condition: odd digits occupy even places

This means all the odd digits $1,1,3$ must be placed in even positions only.

So first, choose $3$ of the $4$ even positions for these odd digits: $\binom{4}{3}=4$

Arrange the odd digits in the chosen even positions

We must arrange $1,1,3$ in those $3$ selected places.

Number of distinct arrangements: $\frac{3!}{2!}=3$

So, total ways to place odd digits: $4 \times 3 = 12$

Arrange the remaining even digits in the remaining places

After placing the odd digits, the remaining digits are: $2,2,2,2,4,4$ These occupy the remaining $6$ positions.

Number of distinct arrangements of these $6$ digits: $\frac{6!}{4!2!} = \frac{720}{24\cdot 2}=15$

Total number of required numbers

Multiply the two independent counts: $12 \times 15 = 180$

Check options

The correct value is: $180$

Hence, the correct option is D.