Let $n$ be the smallest counterexample to the Collatz conjecture.
Theorem: If $n$ exists, then it is odd.
Proof: We will proceed by contradiction and assume that $n$ is even and therefore: $$n = 2p, p \in \mathbb{Z}$$ Then since $n$ is even, the next step of the Collatz sequence is $$\frac{n}{2} = p$$ So $p$ would not satisfy the Collatz conjecture, and be smaller than $n$. Therefore $n$, if it exists, must be odd.
Lets proceed with the knowledge that $n$ is odd. That means the Collatz sequence must proceed: $$\{n,3n+1,\frac{3n+1}{2}\}$$ Since $3n+1$ is necessarily even. We can observe $$\frac{3n+1}{4} < n$$ For all $n > 2$. So we would have another contradiction if $\frac{3n+1}{2}$ is even, and therefore it must be odd.