I'm going to assume your true question is finding an answer that you do not consider "cheating."

Question/Problem
Find a function $f:\mathbb{R}\to\mathbb{R}$ such that $f$ is discontinuous at each point in $K\overset{\text{def}}{=}\{\frac{1}{n}:n\in\mathbb{N}\text{ and }n\ne 0\}\cup\{0\}$ and $f$ is continuous at each point in the complement of $K$ which is denoted $(\mathbb{R}\setminus K)$

General Answer
Let $g:\mathbb{R}\to\mathbb{R}$ be an arbitrary continuous function. Let $\epsilon>0$ be an arbitrary positive real number.

Your edited answer has $g$ be the zero function and $\epsilon$ be $1$

Define $f:\mathbb{R}\to\mathbb{R}$ by $$f(x)=\begin{cases} g(x)+\epsilon&\text{if }x\in K\\ g(x)&\text{if }x\in(\mathbb{R}\setminus K) \end{cases}$$ for every $x\in\mathbb{R}$ where $K\overset{\text{def}}{=}\{\frac{1}{n}:n\in\mathbb{N}\text{ and }n\ne 0\}\cup\{0\}$.

The reason why I introduce "$K$" is because this method can be used for any given set where you want discontinuities. The set you were given is not special in any way for this problem. However it is a classic example of a compact set, but that's not relevant to your problem.

I'm going to assume your true question is finding an answer that you do not consider "cheating."

Question/Problem
Find a function $f:\mathbb{R}\to\mathbb{R}$ such that $f$ is discontinuous at each point in $K\overset{\text{def}}{=}\{\frac{1}{n}:n\in\mathbb{N}\text{ and }n\ne 0\}\cup\{0\}$ and $f$ is continuous at each point in the complement of $K$ which is denoted $(\mathbb{R}\setminus K)$

General Answer
Let $g:\mathbb{R}\to\mathbb{R}$ be an arbitrary continuous function. Let $\epsilon>0$ be an arbitrary positive real number.

Your edited answer has $g$ be the zero function and $\epsilon$ be $1$

Define $f:\mathbb{R}\to\mathbb{R}$ by $$f(x)=\begin{cases} g(x)+\epsilon&\text{if }x\in K\\ g(x)&\text{if }x\in(\mathbb{R}\setminus K) \end{cases}$$ for every $x\in\mathbb{R}$ where $K\overset{\text{def}}{=}\{\frac{1}{n}:n\in\mathbb{N}\text{ and }n\ne 0\}\cup\{0\}$.

The reason why I introduce "$K$" is because this method can be used for any given nowhere dense set $K$ where you want discontinuities. The set you were given is not special in any way for this problem. However it is a classic example of a compact set, but that's not relevant to your problem.

I'm going to assume your true question is finding an answer that you do not consider "cheating."

Question/Problem
Find a function $f:\mathbb{R}\to\mathbb{R}$ such that $f$ is discontinuous at each point in $K\overset{\text{def}}{=}\{\frac{1}{n}:n\in\mathbb{N}\text{ and }n\ne 0\}\cup\{0\}$ and $f$ is continuous at each point in the complement of $K$ which is denoted $(\mathbb{R}\setminus K)$

General Answer
Let $g:\mathbb{R}\to\mathbb{R}$ be an arbitrary continuous function. Let $\epsilon>0$ be an arbitrary positive real number.

Your edited answer has $g$ be the zero function and $\epsilon$ be $1$

Define $f:\mathbb{R}\to\mathbb{R}$ by $$f(x)=\begin{cases} g(x)+\epsilon&\text{if }x\in K\\ g(x)&\text{if }x\in(\mathbb{R}\setminus K) \end{cases}$$ for every $x\in\mathbb{R}$ where $K\overset{\text{def}}{=}\{\frac{1}{n}:n\in\mathbb{N}\text{ and }n\ne 0\}\cup\{0\}$.

I'm going to assume your true question is finding an answer that you do not consider "cheating."

Question/Problem
Find a function $f:\mathbb{R}\to\mathbb{R}$ such that $f$ is discontinuous at each point in $K\overset{\text{def}}{=}\{\frac{1}{n}:n\in\mathbb{N}\text{ and }n\ne 0\}\cup\{0\}$ and $f$ is continuous at each point in the complement of $K$ which is denoted $(\mathbb{R}\setminus K)$

General Answer
Let $g:\mathbb{R}\to\mathbb{R}$ be an arbitrary continuous function. Let $\epsilon>0$ be an arbitrary positive real number.

Your edited answer has $g$ be the zero function and $\epsilon$ be $1$

Define $f:\mathbb{R}\to\mathbb{R}$ by $$f(x)=\begin{cases} g(x)+\epsilon&\text{if }x\in K\\ g(x)&\text{if }x\in(\mathbb{R}\setminus K) \end{cases}$$ for every $x\in\mathbb{R}$ where $K\overset{\text{def}}{=}\{\frac{1}{n}:n\in\mathbb{N}\text{ and }n\ne 0\}\cup\{0\}$.

The reason why I introduce "$K$" is because this method can be used for any given set where you want discontinuities. The set you were given is not special in any way for this problem. However it is a classic example of a compact set, but that's not relevant to your problem.