The statement, as it reads, is true: $\forall n \in \mathbb{Z}^+, \exists a = n \in \mathbb{Z}^+$, such that $a\mid n$, and $\dfrac na = \dfrac nn = 1$ is odd.
IF it is also required that $a \neq n$, then one counterexample suffices to prove the statement is false: for $n = 4$, exists $a=1$ or $a = 2$ or $a = 4$ $\implies \dfrac {n}{a}$ is even. Since $3\not\mid 4,$ and no $a>4$ divides 4.
Hence, the statement, with this modification, is false. And you have thus proved that the negation of the statement is therefore true.
You are right when you say that $\forall x$, the statement $x \in \emptyset$ is false. This means that $\forall x, \neg (x \in \emptyset)$, which is equivalent to $\forall x, x \notin \emptyset$. Perfectly true.
Then you say "the statement $\forall x \in \emptyset$ is false". $\forall x \in \emptyset$ is NOT a statement, it's an incomplete sentence. Either you write "$\forall x, P(x)$", either you write "$\forall x \in X, P(x)$", which is a shorthand for "$\forall x, (x \in X \implies P(x))$". "$\forall x \in \emptyset$" is not a statement. It can't be true or false.
$\forall x \in \emptyset, P(x)$ is a shorthand for $\forall x, (x \in \emptyset \implies P(x))$, which is equivalent (since $x \in \emptyset$ is always false) to $\forall x, ~\textrm{false} \implies P(x)$. After looking at the truth table for $\implies$, this is equivalent to $\forall x, ~\textrm{true}$ (whatever $P(x)$ may be), which is $\textrm{true}\;.$
If you want to disprove $\forall x \in \emptyset, P(x)$ you have to show me an $x \in \emptyset$ such that $P(x)$ is false. Well you will never find an $x \in \emptyset$.
Best Answer
You are correct in saying that that statement $x<y$ alone does not imply the existence of some $z<y$ such that $x<z$. This would indeed be true if we were not given any more information on what the set $\mathbb{Q}$ is.
However, as I'm sure you're aware, $\mathbb{Q}$ (the set of rational numbers) is very much defined, with many of its properties being taken as known assumptions.
One such property of $\mathbb{Q}$ is that it is closed under addition, meaning
$$\forall p\forall q(p,q\in\mathbb{Q}\implies p+q\in\mathbb{Q})$$
Another property is that $\mathbb{Q}$ has multiplicative inverses, except for the cases when dividing by zero:
$$\forall p\forall q(p,q\in\mathbb{Q}\wedge q\neq 0\implies \frac{p}{q}\in\mathbb{Q})$$
With these two properties, we can conclude that if $x,y\in\mathbb{Q}$, then $x+y\in\mathbb{Q}$.
Thus, since $x+y\in\mathbb{Q}$, then $\frac{x+y}{2}\in\mathbb{Q}$. Trivially, when $x<y$, then $x<\frac{x+y}{2}<y$.
Hence there definitely does exist $z$ such that, $x<z<y$, at least for the case where $z=\frac{x+y}{2}$.