Consider two $n$ qubit density matrices:
$$ \rho = \frac{1}{N_a} \left(\sum_{i=1}^{N_a} |\psi_i\rangle \langle \psi_i| \right). $$
$$ \sigma = \frac{1}{N_b} \left(\sum_{i=1}^{N_b} |\phi_i\rangle \langle \phi_i| \right). $$
Now, let $N_a > N_b$.
Is it true that the trace distance
$$ |\rho - \sigma|_1 \geq \frac{N_a - N_b}{N_a}~~? $$
This is not true in general: One can find three pure states $\psi_1,\psi_2,\psi_3$ which, when equally mixed, make up the maximally mixed state, so choosing $\phi_1,\phi_2$ any orthonormal basis of $\mathbb C^2$ yields $\rho-\sigma=0$ while the $\{\phi_j\}_j$ and the $\{\psi_j\}_j$ can be "quite far apart".
But this is by no means the only counterexample — a general, yet somewhat trivial construction goes as follows: Given any $\{\phi_1,\ldots,\phi_{N_B}\}\subset\{x\in\mathbb C^2:\|x\|=1\}$, define $N_A:=2N_B$ and${}^1$ \begin{align*} \psi_1&:=\phi_1=:\psi_2\\ \psi_3&:=\phi_2=:\psi_4\\ &\vdots\\ \psi_{N_A-1}&:=\phi_{N_B}=:\psi_{N_A}\,. \end{align*} Hence $$ \rho=\frac1{N_A}\sum_i|\psi_i\rangle\langle\psi_i|=\frac1{2N_B}\sum_i2|\phi_i\rangle\langle\phi_i|=\frac1{N_B}\sum_i|\phi_i\rangle\langle\phi_i|=\sigma $$ and hence $\|\rho-\sigma\|_1=0<\frac{N_A-N_B}{N_A}$. Even if you assume that $\{\psi_j\}_j\cap\{\phi_j\}_j=\emptyset$ you can just perturb the isometry $V$ with respect to which the $\psi_j$ emerge from the $\phi_j$ and still get $\|\rho-\sigma\|_1=0$. In particular, this works for all $N_A>N_B$.
${}^1$: To connect this to the first example: this prodecure corresponds to the isometry $V:\mathbb C^{N_B}\to\mathbb C^{2N_B}$, $V:={\bf1}\otimes \frac1{\sqrt2}(1,1)^T$
