Partitioning $\beta \mathbb{Z} \setminus \mathbb{Z}$ Take the integers $\mathbb{Z}$ and the addition
\begin{align*}
  +: \mathbb{Z} \times \mathbb{Z} &\to \mathbb{Z}
  \\
   (a,b) &\mapsto a+b.
\end{align*}
Using the Stone-Čech compactification $\beta\mathbb{Z}$ in two steps, $A$ can be "continuously" extended to
\begin{equation*}
  +: \beta\mathbb{Z} \times \mathbb{Z} \to \beta\mathbb{Z}
\end{equation*}
and then to
\begin{equation*}
  +: \beta\mathbb{Z} \times \beta\mathbb{Z} \to \beta\mathbb{Z}.
\end{equation*}
The resulting operation $+$ is continuous in its right argument, and, when the right argument is an integer, it is continuous on its left argument.
Call $\mathbb{Z}^* := \beta\mathbb{Z} \setminus \mathbb{Z}$.
For any $p \in \mathbb{Z}^*$, we have the right ideal $I_p := p + \beta\mathbb{Z}$.
Is it possible to partition $\mathbb{Z}^*$ in two (non trivial) pieces $A$ and $B$ such that
\begin{equation*}
  p \in A \text{ and } q \in B
  \Rightarrow
  I_p \cap I_q = \emptyset?
\end{equation*}
In other words, if I generate an equivalence relation based on $p \sim q$ whenever $I_p \cap I_q \neq \emptyset$, will this equivalence relation partition $\mathbb{Z}^*$ in more then one piece?
 A: Yes, this is possible.
First of all, let me suggest a way of thinking about the $+$ operation. If you're familiar with the idea of taking limits along an ultrafilter, then given $p,q \in \mathbb Z^*$, you can think of $p+q$ as 
$$p+q = q\text{-lim}_{n \in \mathbb Z} \ (p+n).$$
I tend to look at this dynamically: we have a map (the shift map) that sends an ultrafilter $p$ to $p+1$, and $p+q$ just represents one limit point of the set $\{p+n \,:\, n \in \mathbb Z\}$, the (full) orbit of $p$ under the shift map. Also, this makes it clear that $I_p = p+\beta \mathbb Z = \overline{\{p+n \,:\, n \in \mathbb Z\}}$ for any $p \in \mathbb Z^*$.
To answer your question, one partition of $\mathbb Z^*$ that does what you want is the partition into the two clopen sets $\mathbb N^*$ and $(-\mathbb N)^*$ (the ultrafilters concentrating on the positive and negative integers, respectively). If $p$ concentrates on the positive integers, then so does $p+n$ for any $n$. In other words, $p \in \mathbb N^*$ implies $p+n \in \mathbb N^*$ for all $n$. This means that $p+\beta \mathbb Z = \overline{\{p+n \,:\, n \in \mathbb Z\}} \subseteq \mathbb N^*$. Similarly, if $p \in (-\mathbb N)^*$ then $p+\beta \mathbb Z = \overline{\{p+n \,:\, n \in \mathbb Z\}} \subseteq (-\mathbb N)^*$.
No other partition of $\mathbb Z^*$ into clopen sets has the property you want. (I'll sketch a proof of this below.) 
But there are other, more interesting partitions of $\mathbb Z^*$ that do what you want. Recall that $X \subseteq \mathbb Z^*$ is a weak $P$-set if $X \cap \overline{D} = \emptyset$ for any countable $D \subseteq \mathbb Z^* \setminus X$. In this paper, Jonathan Verner and I show that there is a minimal right ideal $R$ of $\mathbb Z^*$ that is also a weak $P$-set.  This means that if $p \in R$, then $p + \beta \mathbb Z = R$ (because it's a minimal right ideal), and if $p \notin R$ then $p+\beta \mathbb Z = \overline{\{p+n \,:\, n \in \mathbb Z\}} \subseteq \mathbb Z^* \setminus R$ (because $R$ and its complement are both shift-map-invariant, and because $R$ is a weak $P$-set). Therefore the partition of $\mathbb Z^*$ into $R$ and $\mathbb Z^* \setminus R$ has the property you want. In fact, the proof in my paper with Jonathan shows that there are $2^{\mathfrak{c}}$ distinct right ideals of this kind; because each minimal right ideal that is a weak $P$-set constitutes an equivalence class, it follows that the equivalence relation you describe has the maximum possible number of equivalence classes, namely $2^{\mathfrak{c}}$.
(Suppose $\mathcal P$ is a partition of $\mathbb Z^*$ into clopen sets other than the one described above. Some $A^* \in \mathcal P$ has the property that either (1) $\mathbb N^* \cap A^* \neq \emptyset$ and $\mathbb N^* \setminus A^* \neq \emptyset$, or else (2) $(-\mathbb N^*) \cap A^* \neq \emptyset$ and $(-\mathbb N^*) \setminus A^* \neq \emptyset$. Without loss of generality, let us say $\mathbb N^* \cap A^* \neq \emptyset$ and $\mathbb N^* \setminus A^* \neq \emptyset$. This means that the set $E$ of $n \in \mathbb N$ such that $n \in A$ but $n+1 \notin A$ must be infinite. If $p$ is an ultrafilter that concentrates on $E$, then we have $p \in A^*$ while $p+1 \notin A^*$. Thus $p$ and $p+1$ are in different partition elements, but clearly $p + \beta \mathbb Z = \overline{\{p+n \,:\, n \in \mathbb Z\}} = \overline{\{(p+1)+n \,:\, n \in \mathbb Z\}} = (p+1)+\mathbb Z^*$. Therefore this partition does not have the property you want.)
