Let $p(t)=\det(A-tI)$ be the characteristic polynomial of the matrix $A$.
It is a degree $n$ polynomial and the coefficients are real numbers since $A$ is a real matrix. Since $n$ is odd, the leading term of $p(t)$ is $-t^n$. That is, we have
\[p(t)=-t^n+\text{lower terms}.\]
(Note: if you use the alternative definition of characteristic polynomial $p(t)=\det(tI-A)$, then the leading term is $t^n$. You can easily modify the proof for this case.)

Therefore, as $t$ increases the polynomial $p(t)$ tends to $-\infty$:
\[\lim_{t\to \infty}=-\infty.\]
Similarly, we have
\[\lim_{t \to -\infty}=\infty.\]
By the intermediate value theorem, there is $-\infty < \lambda < \infty$ such that$p(\lambda)=0$.
Since a root of $p(t)$ is an eigenvalue of $A$, we have obtained a real eigenvalue $\lambda$ of $A$.

Proof 2.

Let $p(t)=\det(A-tI)$ be the characteristic polynomial of the matrix $A$ and write it as
\[p(t)=\prod_{i=1}^k(\lambda_i-t)^{n_i},\]
where $\lambda_i$ are distinct eigenvalues of $A$ and $n_i$ is the algebraic multiplicity of $\lambda_i$.

Since $A$ is a real matrix, the coefficients of $p(t)$ are real numbers. So we have
\[\overline{p(t)}=p(\,\bar{t}\,).\]
Thus, we have
\begin{align*}
p(\bar{t})&=\overline{p(t)}=\prod_{i=1}^k(\,\bar{\lambda}_i-\bar{t}\,)^{n_i}.
\end{align*}
Replacing $\bar{t}$ by $t$, we obtain
\begin{align*}
p(t)=\prod_{i=1}^k(\bar{\lambda}_i-t)^{n_i}.
\end{align*}
This yields that the complex conjugate $\bar{\lambda}_i$ is an eigenvalue with algebraic multiplicity $n_i$.
Hence each non-real eigenvalue $\lambda$ appears as a pair $(\lambda, \bar{\lambda})$ counting multiplicity.
Thus, there are even number of non-real eigenvalues of $A$.

The number of all eigenvalues of $A$ counting multiplicities is $n$.
Since $n$ is odd, there must be at least one real eigenvalue of $A$.

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