All possible numbers are formed using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 taken all at time. The number of such numbers in which the odd digits occupy even places is 

The given digits are 1, 1, 2, 2, 2, 2, 3, 4, 4

There are 9 digits, in this odd digits are $1,1,3$

In 9 places there are 4 even places, these 4 even placess can be filled with $1,1,3$ and arrange among them selves in $\frac{4_{P_3}}{2!}=\frac{4!}{2!}=12$

The remaining 6 blanks ( 3 blanks filled by either 1, 1, or 3) can be filled with the remaining six digits $2,2,2,2,4,4$ in $\frac{6!}{4!2!}=15$ 

Therefore, the number of numbers in which the odd digits occupy even places is $12·15=\boxed{180}$