Number the places from $0$ to $4015$ It is obviously there has to exist an $L$ such that $a+L$ has an other gender sitting on this number than $a$ where we look modulo $4016$. If $gcd(L,4016)=T$, we know the first $T$ places makes all genders will be fixed at all places. claim 1: The configuration is fixed when we know the gender on places $0$ until $T-1$ and correct if $16 \not|T$ If $16|T$ then looking $251T$ places away, we have the other gender, but $4016|251T$ hence $16 \not| T$ As $2| \frac{4016}{T}$ it is easy to see $a+T$ will have the other gender as $a$ does. ( $a+2Lk$ not $\equiv a+T \pmod{4016} $) and exactly half of the people sitting will be men, other half being women. Hence we know the whole fixed thing. claim 2: For different $T_1,T_2$ we won't have twice the same configuration if $v_2(T_1) \not = v_2(T_2)$ Proof: wlog $v_2(T_1)> v_2(T_2)$ hence there exist odd $k$ and even $m$ such that $kT_1=mT_2$ and hence the gender at place $a+kT_1=a+mT_2$ have to be same and different as the gender on place $a$, contradiction. Hence we can't count twice a same configuration. Claim 3 If we have made all configurations for $T_1=251*2^k$ witth $m$ an odd number, the configurations for $T_2=2^k$ are already counted . Proof: As we know $a+T_2$ and $a+252T_2$ have different genders, Taking $b=a+T_2$ , we get on $b$ and $b+T_1$ there are different genders. With knowing which gender are on the first $T_2$ chairs, we know those on $T_1$ first chairs and with these we get exactly the same configuration. MAIN We have to count the numbers of configurations for $T=251,502,1004,2008$ by previous claims which are $2^T$ each time and hence the total will be $2^{251}+2^{502}+2^{1004}+2^{2008}$ ways.