I am going to show that this is false. Take the negation of the original statement $(0)$:
$$\exists x\in\Bbb R\forall y\in\Bbb R\exists z\in\Bbb R xy\ne xz\tag1$$
Now let $x=1$.
$$\forall y\in\Bbb R\exists z\in\Bbb R y\ne z\tag2$$
Clearly we can take $z=y+1$ to make $y\ne z$. $(2)$ is thus true, and since $x=1$ is the element for which this was proved true, $(1)$ is true and $(0)$ is false.
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 $z=\frac{x+y}{2}$.
Best Answer
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.