Let $n$ be a fixed positive integer. Define a relation $R$ on the set $Z$ of integers by $aRb \Leftrightarrow n\left| {a - b} \right|$. Then $R$ is
1. Reflexive
2. Symmetric
3. Transitive
4. Equivalence

Hint: To know whether $R$ is reflexive, symmetric, transitive or equivalence we’ll start with the proof of each term. For reflexive let $a \in N$ and use the given relation to check if it is reflexive or not. Similarly for symmetric and transitive let $\left( {a,b} \right) \in R$and $\left( {a,b,c} \right) \in Z$such that $\left( {a,b} \right) \in R$,$\left( {b,c} \right) \in R$ respectively.

Formula used:
Let $R$ be a relation in set $A$
Reflexive – If $\left( {a,a} \right) \in R$ for every $a \in R$
Symmetric – If $\left( {a,b} \right) \in R$ then $\left( {b,a} \right) \in R$
Transitive – If $\left( {a,b} \right) \in R$ and $\left( {b,c} \right) \in R$ then $\left( {a,c} \right) \in R$

Complete step by step solution: 
Given that,
Relation $R$ on the set $Z$ of integers is $aRb \Leftrightarrow n\left| {a - b} \right|$
Reflexive –
Let, $a \in N$
$a - a = 0 = 0 \times n$
$ \Rightarrow - a$ is divisible by $n$
Thus $\left( {a,a} \right) \in R\forall a \in Z$
$ \Rightarrow R$ is reflexive
Symmetric –
Let, $\left( {a,b} \right) \in R$
$ \Rightarrow \left( {a - b} \right)$ is divisible by $n$
$\therefore \left( {a - b} \right) = np$for some$p \in Z$
$\left( {b - a} \right) = n\left( { - p} \right)$
$ \Rightarrow \left( {b - a} \right)$ is divisible by $n$
$ \Rightarrow \left( {b,a} \right) \in R$ for all $\left( {a,b} \right) \in Z$
$ \Rightarrow R$ is symmetric
Transitive –
Let, $\left( {a,b,c} \right) \in Z$such that $\left( {a,b} \right) \in R$and $\left( {b,c} \right) \in R$
Since $\left( {a,b} \right) \in R$
$ \Rightarrow \left( {a - b} \right)$ is divisible by $n$
$\therefore \left( {a - b} \right) = np$for some$p \in Z - - - - - \left( 1 \right)$
Also, $\left( {b,c} \right) \in R$
$ \Rightarrow \left( {b - c} \right)$ is divisible by $n$
$\therefore \left( {b - c} \right) = nq$for some$q \in Z - - - - - \left( 2 \right)$
Adding equation (1) and (2)
$a - c = n\left( {p + q} \right)$ for some $p + q \in Z$
$ \Rightarrow \left( {a,c} \right) \in R$ for all $\left( {a,b,c} \right) \in Z$
$ \Rightarrow R$ is transitive
Since $R$ is reflexive, symmetric and transitive.
It implies that $R$ is an equivalence relation.
So, Option (4) is the correct answer..

Note: In such question, use the given condition $aRb \Leftrightarrow n\left| {a - b} \right|$ till last to prove all the relations. Here, an equivalence relation is a reflexive, symmetric, and transitive binary connection. To Prove that the relation is equivalence one should prove all the three relations. If any of the relations will not satisfy then it will not be an equivalence relation.