There are two possible interpretations here, the first, permutations
consisting of an even number of odd cycles and some even cycles and
second, permutations consisting of an even number of odd cycles only.
First interpretation.
Observe that the generating function of permutations with odd cycles
marked is
$$G(z, u) =
\exp\left(\sum_{k\ge 1} \frac{z^{2k}}{2k}
+ u \sum_{k\ge 0}\frac{z^{2k+1}}{2k+1}\right).$$
This is
$$G(z, u) =
\exp\left((1-u)\sum_{k\ge 1} \frac{z^{2k}}{2k}
+ u \sum_{k\ge 1}\frac{z^{k}}{k}\right).$$
To get the permutations with an even number of odd cycles use
$$\frac{1}{2} G(z,1)+\frac{1}{2} G(z, -1)$$
which yields
$$\frac{1}{2}\exp\left(\sum_{k\ge 1}\frac{z^{k}}{k}\right)
+ \frac{1}{2} \exp\left(2\sum_{k\ge 1} \frac{z^{2k}}{2k}
- \sum_{k\ge 1}\frac{z^{k}}{k}\right).$$
This simplifies to
$$\frac{1}{2} \frac{1}{1-z}
+ \frac{1}{2} (1-z) \frac{1}{1-z^2}$$
which is
$$\frac{1}{2} \frac{1}{1-z}
+ \frac{1}{2} \frac{1}{1+z}.$$
This simply says that when $n$ is even then there must be an even
number of odd cycles and when $n$ is odd there cannot be an even
number of odd cycles, which follows by inspection (parity).
Second interpretation.
Here we have
$$G(z, u) =
\exp\left(u \sum_{k\ge 0}\frac{z^{2k+1}}{2k+1}\right).$$
This is
$$G(z, u) =
\exp\left(u \sum_{k\ge 0}\frac{z^{k}}{k}
- u \sum_{k\ge 1}\frac{z^{2k}}{2k}\right).$$
To get the permutations with an even number of odd cycles use
$$\frac{1}{2} G(z,1)+\frac{1}{2} G(z, -1)$$
which yields
$$\frac{1}{2}\frac{1}{1-z}
\left(\frac{1}{1-z^2}\right)^{-1/2}
+ \frac{1}{2} (1-z)
\left(\frac{1}{1-z^2}\right)^{1/2}.$$
This gives the sequence
$$0, 1, 0, 9, 0, 225, 0, 11025, 0, 893025, 0,
108056025, 0, 18261468225,\ldots$$
which points us to
OEIS A177145,
where we find this computation confirmed.
This generating function maybe written as
$$\frac{1}{2}\frac{1}{1-z}
\sqrt{1-z^2}
+ \frac{1}{2} (1-z)
\frac{1}{\sqrt{1-z^2}}$$
or
$$\frac{1}{2}\sqrt{\frac{1+z}{1-z}}
+ \frac{1}{2}\sqrt{\frac{1-z}{1+z}}.$$
I have started from the stage where I got stuck in proving the above lemma. It is easy to show what I just mentioned in the edit is that $\text {Ord}\ (ab)\ \big |\ \text {lcm}\ \left (\text {Ord}\ (a), \text {Ord}\ (b) \right ).$ To prove equality we need to prove the other way round which is not true for arbitrary finite groups even if $a$ and $b$ commute. We are so lucky that the other part is true for our case. Why? Lets discuss.
Before proving the required result I noticed that if we can prove the following lemma we are through.
Lemma $:$ Let $\sigma, \tau \in S_n$ be two disjoint cycles. Then $\text {Ord}\ (\sigma \tau ) = \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ).$
For proving the equality in the lemma let us first introduce the following definition.
Let $\rho = (a_1,a_2, \cdots , a_r) \in S_n$ be an $r$-cycle. Then the support of $\rho$ is denoted by $\text {Supp}\ (\rho)$ and it is defined as $\text {Supp}\ (\rho) = \{a_1,a_2, \cdots , a_r \}.$ So $\text {Supp}\ (\rho)$ consists of those points in $\{1,2, \cdots, n \}$ which are disturbed by the operation of $\rho.$
Observation $:$ If $\rho,\rho' \in S_n$ are two cycles inverses of each other then $\text {Supp}\ (\rho) = \text {Supp}\ (\rho').$ (Because inverse cycles fix same points).
Now let us take two disjoint cycles $\sigma , \tau \in S_n.$ On contrary let us assume that $\text {Ord}\ (\sigma \tau) = m < \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ).$ Then it is easy to see that $m\ \bigg |\ \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ).$ Let us assume that $\sigma^m \neq \text {id}$ and $\tau^m \neq \text {id}$ for otherwise $m = \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ),$ a contradiction to our assumption. Since fixed points of $\sigma$ and $\tau$ are respectively fixed points of $\sigma^m$ and $\tau^m$ respectively it follows that $\text {Supp}\ (\sigma^m) \subseteq \text {Supp}\ (\sigma)$ and $\text {Supp}\ (\tau^m) \subseteq \text {Supp}\ (\tau).$ Since $\sigma$ and $\tau$ are disjoint cycles so we have $\text {Supp}\ (\sigma) \cap \text {Supp}\ (\tau) = \varnothing.$ Hence $\text {Supp}\ (\sigma^m) \cap \text {Supp}\ (\tau^m) = \varnothing.\ \ \ \ (*)$
Now since $\text {Ord}\ (\sigma \tau) = m$ so we have $$\begin{align*} (\sigma \tau)^m & = \text {id} \implies \sigma^m \tau^m = \text {id} \implies \sigma^m = (\tau^m)^{-1} \end{align*}$$
So $\sigma^m$ is the inverse of $\tau^m.$ So from our Observation it follows that $\text {Supp}\ (\sigma^m) = \text {Supp}\ (\tau^m).$ Since $\sigma^m \neq \text {id}$ and $\tau^m \neq \text {id}$ it follows that $\text {Supp}\ (\sigma^m) = \text {Supp}\ (\tau^m) \neq \varnothing$ and hence $\text {Supp}\ (\sigma^m) \cap \text {Supp}\ (\tau^m) \neq \varnothing,$ which contradicts $(*).$ That implies either $\sigma^m = \text {id}$ or $\tau^m = \text {id}.$ But if one of $\sigma^m$ or $\tau^m$ is identity then by using the equation $\sigma^m \tau^m = \text {id}$ we find that the other is also an identity. So we must have $\sigma^m = \tau^m = \text {id}.$ This implies $\text {Ord}\ (\sigma)\ \big |\ m$ and $\text {Ord}\ (\tau)\ \big |\ m.$ But it means that $\text {lcm}\ \left ( \text {Ord}\ (\sigma),\text {Ord}\ (\tau) \right )\ \bigg |\ m,$ which is a contradiction to our assumption that $m < \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ).$ Hence our assumption is false. So $m \geq \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ).$ But since $m\ \bigg |\ \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right )$ it follows that $m \leq \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ).$ Hence combining these two inequalities it follows that $m = \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ).$
QED
Best Answer
Imagine $k$ subsequences of length $r$ representing the disjoint cycles which you fill up one element at a time with elements of the set $S_n$ acts on. There are $n(n-1)\cdots(n-kr+1)$ ways to do so. The division by $k!$ is because the subsequences may be permuted among themselves; the division by $r^k$ is because cyclic rotations do not change a cycle.