Solution:

The answer is (B). The number of $\pi$ cchi after 7 broods can take any odd value between 7 and $2^{9}-1$, and no other. The answer to the question is therefore $\frac{2^{9}-1-7}{2}+1=2^{8}-3=253$, since this is the number of odd numbers between 7 and $2^{9}-1=511$.

We justify the previous statement through the following facts:

a. a family consisting of only 1 $\pi$ cchio will forever continue to consist of a single $\pi$ cchio: obvious;

b. a family consisting of $2 \pi$ chi can give rise, after $n \geq 1$ broods, to no more than $2^{n+1}$ $\pi$ cchi;

c. a family consisting of $3 \pi$ chi can give rise, after $n \geq 1$ broods, to any number of $\pi$ cchi that is odd and between 7 and $2^{n+2}-1$.

We prove statements (b) and (c) simultaneously by induction: assuming that (b) and (c) hold for a certain $n$ we will prove them for $n+1$. Note that the base case $n=1$ is obvious for both. Note also that the parity of the number of $\pi$ chi does not change after a brood, so the parity condition stated in point (c) is indeed necessary. Moreover:

b. in this case, after one brood we have $4 \pi$ cchi, which can live in four families of 1 (in which case at all subsequent generations we will still have $4 \pi$ chi), a family of 2 and two of 1, two families of 2, or a family of 3 and one of 1. In the various cases, statements (b) and (c) for $n$ broods show that - after a further $n$ broods - we will have at most
$$1+1+1+1, \quad 2^{n+1}+1+1, \quad 2^{n+1}+2^{n+1}, \quad\left(2^{n+2}-1\right)+1$$
$\pi$ cchi. Since each of these numbers is less than or equal to $2^{n+2}$, this proves statement (b) for $n+1$ broods.

c. in this case, after one brood the $7 \pi$ cchi can be distributed into families in the following ways:
$$\begin{gathered} 3+3+1, \quad 3+2+2, \quad 3+2+1+1, \\ 3+1+1+1+1, \quad 2+2+2+1, \quad 2+2+1+1+1, \\ 2+1+1+1+1+1, \quad 1+1+1+1+1+1+1 . \end{gathered}$$
After a further $n$ broods, using the inductive hypothesis as above we obtain that in the various cases the total number of $\pi$ cchi is at most
$$\begin{gathered} 2\left(2^{n+2}-1\right)+1, \quad\left(2^{n+2}-1\right)+2\left(2^{n+1}\right), \quad\left(2^{n+2}-1\right)+\left(2^{n+1}\right)+1+1, \\ \left(2^{n+2}-1\right)+1+1+1+1, \quad 3 \cdot\left(2^{n+1}\right)+1, \quad 2 \cdot\left(2^{n+1}\right)+1+1+1 \\ \left(2^{n+1}\right)+1+1+1+1+1, \quad 1+1+1+1+1+1+1 . \end{gathered}$$
It is easy to check that each of these numbers is less than or equal to $2^{n+3}-1$. Finally, if after the first brood the $7 \pi$ chi are organized into families as $3+3+1$, then after a further $n$ broods the number of $\pi$ cchi will be of the form $d_{1}+d_{2}+1$, where $d_{1}, d_{2}$ can take any odd value between 7 and $2^{n+2}-1$. The expression $d_{1}+d_{2}+1$ can then take any odd value between $7+7+1=15$ and $2 \cdot 2^{n+2}-2+1=2^{n+3}-1$. It remains only to show that the number of $\pi$ cchi after $n+1 \geq 2$ broods can also be 7, 9, 11 or 13, but this is easy. Indeed, if at some point the $\pi$ cchi organize into families of 1, from that point on their number will no longer grow, so it suffices to show that these numbers of $\pi$ cchi are achievable after at most two broods, for example as follows:
$$\begin{gathered} 3 \rightarrow 7 ; \quad 3 \rightarrow 2+1+1+1+1+1 \rightarrow 4+1+1+1+1+1=9 \\ 3 \rightarrow 2+2+1+1+1 \rightarrow 4+4+1+1+1=11 ; \quad 3 \rightarrow 2+2+2+1 \rightarrow 4+4+4+1=13 \end{gathered}$$