[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 $m\le 2n$.[/step]

[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: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]