jmc

By the binomial theorem, $(m+n{)}^6$ expands as $$\begin{array}{rl}\left(\genfrac{}{}{0ex}{}{6}{0}\right)m^6+\left(\genfrac{}{}{0ex}{}{6}{1}\right)m^{5}n& +\left(\genfrac{}{}{0ex}{}{6}{2}\right)m^4{n}^2+\left(\genfrac{}{}{0ex}{}{6}{3}\right)m^3{n}^3\\ & +\left(\genfrac{}{}{0ex}{}{6}{4}\right)m^2{n}^4+\left(\genfrac{}{}{0ex}{}{6}{5}\right)m{n}^5+\left(\genfrac{}{}{0ex}{}{6}{6}\right)n^6.\end{array}$$ Since $m$ and $n$ are odd, each of these terms is odd if and only if the binomial coefficient is odd. Since $\left(\genfrac{}{}{0ex}{}{6}{0}\right)=\left(\genfrac{}{}{0ex}{}{6}{6}\right)=1$, $\left(\genfrac{}{}{0ex}{}{6}{1}\right)=\left(\genfrac{}{}{0ex}{}{6}{5}\right)=6$, $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=\left(\genfrac{}{}{0ex}{}{6}{4}\right)=15$, and $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$, exactly $\boxed{4}$ of these terms are odd.