Suppose $n$ is a natural number and $\lambda$ is an unordered integer partition of $n$ such that $\lambda$ has $a_{\lambda,j}$ parts of size $j$ for each $j$... Let $c_\lambda$ be the conjugacy class in the symmetric group of degree $n$ comprising the elements whose cycle type is $\lambda$... Then: $$ \! |c_\lambda| = \frac{n!}{\prod_j (j)^{a_{\lambda,j}}(a_{\lambda,j}!)} $$ (see here for more information: Conjugacy class size formula in symmetric group)
Let $\Lambda$ be the set of all partitions $\lambda$ and $M\subset\Lambda $. For $n\ge5$ there seem to be more than one solution $M$ for the following: $$ \sum_{\mu\in M} |c_\mu|=\frac{n!}2 \tag{1} $$
There is at least one solution, since the splitting into odd and even permutations divides the set of all permutations properly.
For $n=5$ the even/odd splitting looks like $$ \begin{eqnarray} |c_{(1,1,1,1,1)}|+|c_{(3,1,1)}|+|c_{(2,2,1)}|+|c_{(5)}|&=&1+20+15+24\\ |c_{(2,1,1,1)}|+|c_{(4,1)}|+|c_{(3,2)}|&=&10+30+20, \end{eqnarray} $$ but there are at least $3$ more solutions:
Since $|c_{3,2}|=\frac{5!}{(3^1\cdot 1!)(2^1\cdot 1!)}=|c_{3,1,1}|=\frac{5!}{(3^1\cdot 1!)(1^2\dot 2!)}$ you can interchange the odd conjugacy class of the partition $3+2$ with the even one for the partition $3+1+1$.
and 3. $|c_{(1,1,1,1,1)}|+|c_{(2,2,1)}|+|c_{(5)}|=1+15+24=40$, so we may again add $|c_{(3,2)}|$ or $|c_{(3,1,1)}|$
My question:
Is it possible to give an asymptotic? How many solutions $M$ exist for $(1)$ in general?
Ok, it was easier than I thought to find a concise way to calculate the number of solutions:
We've got to calculate $$ \prod_{\lambda\in\Lambda} \left(1+x^{|c_\lambda|}\right) $$
and then find the coefficient of $\displaystyle x^\frac{n!}{2}$. Some results:
$$(1+x)(1+x^{10})(1+x^{20})^2 (1+x^{15})(1+x^{30})(1+x^{24})=x^{120}+x^{119}+x^{110}+x^{109}+x^{105}+x^{104}+2 x^{100}+2 x^{99}+x^{96}+2 x^{95}+x^{94}+3 x^{90}+3 x^{89}+x^{86}+3 x^{85}+2 x^{84}+x^{81}+3 x^{80}+2 x^{79}+2 x^{76}+5 x^{75}+3 x^{74}+x^{71}+4 x^{70}+3 x^{69}+3 x^{66}+5 x^{65}+2 x^{64}+2 x^{61}+\color{red}{4 x^{60}}+2 x^{59}+2 x^{56}+5 x^{55}+3 x^{54}+3 x^{51}+4 x^{50}+x^{49}+3 x^{46}+5 x^{45}+2 x^{44}+2 x^{41}+3 x^{40}+x^{39}+2 x^{36}+3 x^{35}+x^{34}+3 x^{31}+3 x^{30}+x^{26}+2 x^{25}+x^{24}+2 x^{21}+2 x^{20}+x^{16}+x^{15}+x^{11}+x^{10}+x+1$$
$n=6$: $$ (1+x^{1})(1+x^{15})(1+x^{40})^2(1+x^{90})^2(1+x^{45})(1+x^{144})(1+x^{120})^2(1+x^{15})\\=\dots+ \color{red}{12 x^{360}}+\dots $$
$n=7$: $(1+x) \dots (1+x^{840}) = \dots+ \color{red}{58 x^{2520}}+\dots$
$n=8$: $(1+x) \dots (1+x^{5760}) = \dots+ \color{red}{1016 x^{20160}}+\dots$
So summarizing this we get: $0,2,2,2,4,12,58,1016$ (thanks to Jack) for the first eight symmetric groups. The question about the asymptotics remains open...