[proofplan]
Set $m=2k+1$ and use the [Hodge decomposition](/theorems/2745) of the complex cohomology of a compact Kähler manifold. In degree $m$, the summands $H^{p,q}(X)$ satisfy $p+q=m$, so when $m$ is odd no summand lies on the diagonal $p=q$. Hodge symmetry pairs each summand $H^{p,q}(X)$ with $H^{q,p}(X)$ with equal complex dimension, so the total dimension is a sum of doubled integers. In degrees above the real dimension of $X$, the [Hodge decomposition](/theorems/3941) is an empty direct sum, hence the Betti number is $0$.
[/proofplan]
[step:Fix the odd degree and apply the Hodge decomposition]
Fix an integer $k\ge 0$, and define the odd integer $m:=2k+1$. Let $n:=\dim_{\mathbb C}X$. For each pair of integers $p,q\ge 0$, define the Hodge number
\begin{align*}
h^{p,q}(X):=\dim_{\mathbb C}H^{p,q}(X).
\end{align*}
By [citetheorem:8066], the compact Kähler manifold $X$ has a Hodge decomposition in degree $m$:
\begin{align*}
H^m(X;\mathbb C)=\bigoplus_{p+q=m}H^{p,q}(X).
\end{align*}
The summands with $p>n$ or $q>n$ are zero, since $X$ has complex dimension $n$. Therefore
\begin{align*}
b_m(X)=\dim_{\mathbb C}H^m(X;\mathbb C)=\sum_{\substack{p+q=m}}h^{p,q}(X).
\end{align*}
If $m>2n$, there are no pairs $p,q$ with $0\le p,q\le n$ and $p+q=m$, so the sum is empty and $b_m(X)=0$. This is even. Hence it remains to consider the case $m\le 2n$.
[guided]
We fix the degree in which we want to compute the Betti number. Let
\begin{align*}
m:=2k+1.
\end{align*}
This number is odd by construction. Let $n:=\dim_{\mathbb C}X$, so the real dimension of $X$ is $2n$.
For each pair of integers $p,q\ge 0$, define
\begin{align*}
h^{p,q}(X):=\dim_{\mathbb C}H^{p,q}(X).
\end{align*}
These are the Hodge numbers of $X$.
We apply [citetheorem:8066]. Its hypotheses are satisfied because $X$ is compact Kähler by assumption. The theorem gives the Hodge decomposition
\begin{align*}
H^m(X;\mathbb C)=\bigoplus_{p+q=m}H^{p,q}(X).
\end{align*}
Taking complex dimensions of this finite direct sum gives
\begin{align*}
\dim_{\mathbb C}H^m(X;\mathbb C)=\sum_{\substack{p+q=m}} \dim_{\mathbb C}H^{p,q}(X).
\end{align*}
Using the definition of $b_m(X)$ and the definition of $h^{p,q}(X)$, this becomes
\begin{align*}
b_m(X)=\sum_{\substack{p+q=m}}h^{p,q}(X).
\end{align*}
There is one boundary case to isolate. Since $X$ has complex dimension $n$, the only possible nonzero Hodge summands have $0\le p,q\le n$. If $m>2n$, no such pair can satisfy $p+q=m$. Thus the Hodge decomposition in degree $m$ is empty, so
\begin{align*}
H^m(X;\mathbb C)=0.
\end{align*}
Consequently
\begin{align*}
b_m(X)=0.
\end{align*}
This number is even. Therefore the only remaining case is $m\le 2n$.
[/guided]
[/step]
[step:Pair every Hodge summand with its conjugate bidegree]
Assume now that $m\le 2n$. Define the finite index set
\begin{align*}
I_m:=\{(p,q)\in \mathbb Z_{\ge 0}^2: p+q=m\}.
\end{align*}
The involution
\begin{align*}
\sigma:I_m&\to I_m
\end{align*}
is defined by
\begin{align*}
\sigma(p,q):=(q,p).
\end{align*}
Because $m$ is odd, no element of $I_m$ is fixed by $\sigma$: if $(p,q)=\sigma(p,q)$, then $p=q$, hence $m=p+q=2p$, contradicting that $m$ is odd.
By [citetheorem:8081], Hodge symmetry gives
\begin{align*}
h^{p,q}(X)=h^{q,p}(X)
\end{align*}
for every $(p,q)\in I_m$. Choose one representative from each two-element orbit of $\sigma$, and let $R_m\subset I_m$ denote the resulting set of representatives. Then
\begin{align*}
I_m=\bigsqcup_{(p,q)\in R_m}\{(p,q),(q,p)\}.
\end{align*}
Therefore
\begin{align*}
b_m(X)=\sum_{(p,q)\in R_m}\left(h^{p,q}(X)+h^{q,p}(X)\right).
\end{align*}
Using Hodge symmetry in each summand,
\begin{align*}
b_m(X)=\sum_{(p,q)\in R_m}2h^{p,q}(X).
\end{align*}
Thus $b_m(X)$ is divisible by $2$.
[/step]
[step:Translate back to the stated odd Betti number]
Since $m=2k+1$, the preceding steps prove that
\begin{align*}
b_{2k+1}(X)=b_m(X)
\end{align*}
is even. The integer $k\ge 0$ was arbitrary, so $b_{2k+1}(X)$ is even for every integer $k\ge 0$.
[/step]
