1. Initial Setup and Sequence Definition:
The sequence $a_n$ is defined by:
$$a_0 = 1$$
and for $n \geq 0$,
$$a_{n+1} = \begin{cases} \frac{a_n}{2} & \text{if } a_n \text{ is even} \\ a_n + h & \text{if } a_n \text{ is odd} \end{cases}$$

2. **Behavior Analysis for Even $h$:**
If $h$ is even, we need to check if there exists an $n > 0$ such that $a_n = 1$.
- If $a_n$ is even, then $a_{n+1} = \frac{a_n}{2}$.
- If $a_n$ is odd, then $a_{n+1} = a_n + h$.

Since $h$ is even, adding $h$ to an odd number results in an odd number. Thus, $a_n$ will never become 1 if it is odd and $h$ is even. Therefore, for even $h$, there is no $n > 0$ such that $a_n = 1$.

3. **Behavior Analysis for Odd $h$:**
If $h$ is odd, we need to check if there exists an $n > 0$ such that $a_n = 1$.
- If $a_n$ is even, then $a_{n+1} = \frac{a_n}{2}$.
- If $a_n$ is odd, then $a_{n+1} = a_n + h$.

Since $h$ is odd, adding $h$ to an odd number results in an even number. This means that the sequence will alternate between even and odd numbers.

4. **Modulo $h$ Analysis:**
We analyze the sequence modulo $h$. Notice that:
$$a_{n+1} \equiv \begin{cases} \frac{a_n}{2} \pmod{h} & \text{if } a_n \text{ is even} \\ a_n + h \equiv a_n \pmod{h} & \text{if } a_n \text{ is odd} \end{cases}$$

Since $a_0 = 1$, we have $a_0 \equiv 1 \pmod{h}$. The sequence will eventually cycle through values modulo $h$. By Euler's theorem, since $a_n$ is always less than $2h$, there will be a time when $a_i \equiv 1 \pmod{h}$.

5. Conclusion:
For odd $h$, the sequence will eventually reach $a_n = 1$ for some $n > 0$. For even $h$, the sequence will never reach $a_n = 1$.

The final answer is $\boxed{ h }$ must be odd.