We prove the general case where there are $p$ outgoing buses and $q$ returning buses with $p > q$. The number of such tourists is $p - q + 1$. Let $T$ be the set of all tourists. For each $t \in T$, let $p_t$ be the number of tourists in the bus that $t$ takes in the outgoing trip and let $q_t$ be the number of tourists in the bus that $t$ takes in the return trip. Then
$$\sum_{t \in T} \frac{1}{p_t} = p, \quad \sum_{t \in T} \frac{1}{q_t} = q \quad \Rightarrow \quad \sum_{t \in T} \left( \frac{1}{p_t} - \frac{1}{q_t} \right) = p - q$$
Since $|\frac{1}{p_t} - \frac{1}{q_t}| < 1$, at least $p-q+1$ terms in the above sum is positive. That is, there are at least $p-q+1$ tourists such that $\frac{1}{p_t} - \frac{1}{q_t} > 0$, or equivalently $p_t < q_t$.