Yes. If there is a model $\mathcal M$ of size $n$ of any sentence $\phi$ that does not use = then you can take any element $a$ of $\mathcal M$ (using the fact that $n>0$) and let $\mathcal N$ be $\mathcal M$ with an additional element $b$ which has all the same properties as $a$. Then $\mathcal N$ still satisfies $\phi$.

Yes. If there is a model $\mathcal M$ of size $n$ of any sentence $\phi$ that does not use = then you can take any element $a$ of $\mathcal M$ (using the fact that $n>0$) and let $\mathcal N$ be $\mathcal M$ with an additional element $b$ which has all the same properties as $a$. Then $\mathcal N$ still satisfies $\phi$.

If $n=0$, however, the answer is no: the sentence $$\exists x(P(x)\vee\neg P(x))\longrightarrow \exists x\exists y(P(x)\wedge\neg P(y))$$ has a model of size 0 but no model of size 1.