, 2020

Solution:
Define
$$\mathbf{1}_{a \mid n}= \begin{cases}1 & a \mid n \\ 0 & \text{otherwise}\end{cases}$$
Then
$$\begin{aligned} f(n)^2 & =\left(\mathbf{1}_{2 \mid n}+\mathbf{1}_{3 \mid n}+\mathbf{1}_{5 \mid n}\right)^2 \\ & =\mathbf{1}_{2 \mid n}+\mathbf{1}_{3 \mid n}+\mathbf{1}_{5 \mid n}+2\left(\mathbf{1}_{2 \mid n} \mathbf{1}_{3 \mid n}+\mathbf{1}_{2 \mid n} \mathbf{1}_{5 \mid n}+\mathbf{1}_{3 \mid n} \mathbf{1}_{5 \mid n}\right) \\ & =\mathbf{1}_{2 \mid n}+\mathbf{1}_{3 \mid n}+\mathbf{1}_{5 \mid n}+2\left(\mathbf{1}_{6 \mid n}+\mathbf{1}_{10 \mid n}+\mathbf{1}_{15 \mid n}\right) \end{aligned}$$
So summing $f(n)^2$ over integers $1 \leq n \leq 2020$ is the same as summing $1$ for each time $n$ is divisible by $2$, $3$, or $5$, and additionally summing $2$ for each time $n$ is divisible by $6$, $10$, or $15$.
$$$$
$$\begin{aligned} \sum_{n=1}^{2020} f(n)^2 & =\left\lfloor\frac{2020}{2}\right\rfloor+\left\lfloor\frac{2020}{3}\right\rfloor+\left\lfloor\frac{2020}{5}\right\rfloor+2\left(\left\lfloor\frac{2020}{6}\right\rfloor+\left\lfloor\frac{2020}{10}\right\rfloor+\left\lfloor\frac{2020}{15}\right\rfloor\right) \\ & =1010+673+404+2(336+202+134)=3431 \end{aligned}$$