With the usual order $I=\mathbb N$ is a directed set.
Suppose that for each $i\in \mathbb N$ we have a subset $\{s_i\}$ of real numbers, such that for each $i\leq j$, $s_i\mathbb Z$ is a subgroup of $s_j\mathbb Z$ and the inclusion homomorphism are the transtition maps $\varphi_{ij}:s_i\mathbb Z\rightarrow s_j\mathbb Z$.
Then the direct limit is subgroup of $\mathbb Z$ right? because each $s_i \mathbb Z$ is a subgroup of $\mathbb Z$ and the direct limit is identified with a subgroup of $\mathbb Z$.
Note that for real numbers $a$ and $b$, if $a\mathbb{Z}\leq b\mathbb{Z}$ then $a\in b\mathbb{Z}$, so there exists an integer $n$ such that $a=nb$; if $a\neq 0$, then this means that $b=\frac{1}{n}a$.
Any $0$ term forces all previous terms to be $0$; if all terms are $0$, you just get $\{0\}$. If at least one term is nonzero, we may assume that it is $s_1$. So assume $s_1\neq 0$.
Then there must exist integers $n_1,n_2,\ldots$ such that $$\begin{align*} s_2 &= \frac{1}{n_1}s_1\\ s_3 &= \frac{1}{n_2}s_2 = \frac{1}{n_1n_2}s_1\\ &\vdots \end{align*}$$ Dividing through by $s_1$ we may assume that what you have is $$\mathbb{Z}\supseteq\frac{1}{n_1}\mathbb{Z}\supseteq\frac{1}{n_1n_2}\mathbb{Z}\supseteq\cdots \supseteq \frac{1}{n_1\cdots n_k}\mathbb{Z}\supseteq\cdots$$
The colimit/union that you get is: $$\left\{\frac{a}{b}\in\mathbb{Q}\;\Bigm|\; \gcd(a,b)=1\text{ and there exists }k\text{ such that }b\mid n_1\cdots n_k\right\}$$
(Okay, "really" it's the set of all $\frac{s_1a}{b}$ with $\gcd(a,b)=1$, etc. but we may as well take $s_1=1$ to look at the structure...)
Indeed, if $\frac{a}{b}$ is as given, then let $c$ be such that $bc=n_1\cdots n_k$. Then $\frac{a}{b}= \frac{ac}{n_1\cdots n_k}\in \frac{1}{n_1\cdots n_k}\mathbb{Z}$.
Conversely, every element of $\bigcup_{k=1}^{\infty}\frac{1}{n_1\cdots n_k}\mathbb{Z}$ is rational; if $\frac{a}{b}$ lies in the union, with $\gcd(a,b)=1$, then there exists $k$ and an integer $m$ such that $\frac{a}{b}=\frac{m}{n_1\cdots n_k}$. This gives that $b\mid n_1\cdots n_k$, as desired.
In particular, if the sequence $|n_1|,|n_2|,\ldots$ is eventually constant $1$, you get a group isomorphic to $\mathbb{Z}$, since you get $\frac{1}{N}\mathbb{Z}$ for some integer $N\gt 1$.
If the sequence is not eventually constant $1$, then the resulting group cannot be isomorphic to $\mathbb{Z}$, as it contains fractions with arbitrarily large denominator, hence it cannot be a cyclic subgroup of $\mathbb{Q}$.