The popcorn function

Rational points. Let $a=\frac pq$ be rational. Then $f(a)=\frac1q>0$. Every neighbourhood of $a$ contains irrational numbers, where $f=0$, so $f$ cannot tend to $f(a)>0$. Hence $f$ is discontinuous at $a$.

Irrational points. Let $a$ be irrational and $\varepsilon>0$. Choose an integer $N>\frac1\varepsilon$.

*Finitely many bad rationals.* Consider the rationals in $[a-1,a+1]$ whose denominator (in lowest terms) is at most $N$. For each fixed denominator $q\le N$ there are only finitely many such fractions $\frac pq$ in a bounded interval, so the set $B$ of all of them is finite. Since $a$ is irrational, $a\notin B$, so
$$\delta=\min\Bigl(1,\ \min_{b\in B}|a-b|\Bigr)>0.$$

*Checking.* Let $|x-a|<\delta$. If $x$ is irrational, $f(x)=0$. If $x=\frac pq$ is rational, then $x\notin B$, so $q>N$ and $f(x)=\frac1q<\frac1N<\varepsilon$. In both cases $|f(x)-f(a)|=f(x)<\varepsilon$.

Conclusion. $f$ is continuous exactly at the irrational numbers.

Remark. A function can be continuous at infinitely many points and discontinuous at infinitely many others, with both sets dense. The graph looks like popcorn: tall kernels $\frac12,\frac13,\dots$ above each rational, flattening toward $0$. The reverse is impossible: there is no function continuous exactly at the rationals.