$\textbf{Motivation:}$ Van der Vaart - Asymptotic statistics [Example 5.11]
$\textbf{Setup:}$ Suppose I have $i.i.d$ random variables $X_1,...,X_n$. Define the sample median $\hat\theta_n(\omega)$ as the median value of $\{X_1(\omega),...,X_n(\omega)\}$ for every $\omega \in \Omega$. Define the estimator $\theta \mapsto \Psi_n(\theta) \equiv \frac{1}{n}\sum_{i=1}^nsign(X_i-\theta)$.
$\textbf{Question:}$ How do I show that $\Psi_n(\hat\theta_n) = o_p(1)?$
I know we can first write $\Psi_n(\hat\theta_n) = \frac{1}{n}\sum_{i=1}^n (1\{ X_i - \hat\theta_n >0 \} - 1\{X_i - \hat\theta_n<0 \} ) $, but it seems that writing out $\hat\theta_n$ mathematically is challenging.
If $\hat \theta_n$ is a sample median, then at least $\lfloor n/2\rfloor$ of the $X_i$ are on each side of $\hat\theta_n$. So (as you have in your question):
$$|\Psi_n(\hat \theta_n)| = \frac{1}{n}\left|\sum_{i=1}^n 1\{ X_i > \hat\theta_n \} - 1\{X_i < \hat\theta_n \} \right| \leq \frac{1}{n} $$
Since the quantity inside the absolute value is $0$ when $n$ is even and there are no ties, and cannot be larger than $1$ since otherwise $\hat\theta_n$ is not a sample median.