This problem is proposed by me and Professor Yang. We need a well-know result: $$\sum_{p \text{ is prime }}\frac{1}{p^2}<0.48.$$ Now back to our problem: we try to find $x\in \mathbb{Z}_+$, such that $x$ and $x+2020$ are all square-free, but $x+i$ are not square-free($1\leq i\leq 2019$). We select $2019$ distinct $p_1,p_2,\cdots,p_{2019}$, $p_i>4039(1\leq i\leq 2019)$. Select $x_k=k\left(\prod_{i=1}^{2019}p_i\right)^2+x$, where $k\in \mathbb{Z}_+$, and $x$ satisfies: \[ \left\{ \begin{aligned} x+1 &\equiv 0 \pmod{p_1^2} \\ x+2 &\equiv 0 \pmod{p_2^2} \\ &\cdots \\ x+2019 &\equiv 0 \pmod{p_{2019}^2} \end{aligned} \right. \]CRT tell us it's Okay. Now we hope to find $k\in \mathbb{Z}_+$, such that $x_k,x_k+2020$ are not square-free: We fix a large enough $N$, $N>2+40000\left(\prod_{i=1}^{2019}p_i\right)^2$, then $0.01N>2\left(\prod_{i=1}^{2019}p_i\right)\sqrt{N+2}$. $p_i>4039(1\leq i\leq 2019)$ tell us $p_i \nmid x_k, x_k+2020$($1\leq i\leq 2019$). For any prime $p$, $k\left(\prod_{i=1}^{2019}p_i\right)^2+x\equiv 0\pmod{p^2}$ has at most $\left\lceil \frac{N}{p^2} \right\rceil$ solutions when $1\leq k\leq N$; and $k\left(\prod_{i=1}^{2019}p_i\right)^2+x+2020\equiv 0\pmod{p^2}$ also has at most $\left\lceil \frac{N}{p^2} \right\rceil$ solutions when $1\leq k\leq N$ Let $\prod_{i=1}^{2019}p_i=M$, if $p^2\mid kM^2+x$ or $p^2\mid kM^2+x+2020$, then $p\leq \sqrt{kM^2+2M^2}\leq M\sqrt{N+2}$. Now, $$N-2\sum_{\substack{p \text{ is prime }\\ p\leq M\sqrt{N+2}}}\left\lceil \frac{N}{p^2} \right\rceil\geq N-2N\sum_{p\text{ is prime }}\frac{1}{p^2}-2M\sqrt{N+2}>0.04N-2M\sqrt{N+2}\rightarrow +\infty, N \rightarrow +\infty.$$ We have done.

Solved with Th3Numb3rThr33. To begin, let \(p_1\), \(p_2\), \ldots, \(p_{2019}\) be primes, with \(P=p_1p_2\cdots p_{2019}\), and let \(c\) be an integer (that exists by Chinese Remainder theorem) such that \(c\equiv -i\pmod{p_i^2}\) for \(i=1,\ldots,2019\). Then for each \(k\) of the form \(k=P^2i+c\), the integers \(k+1\), \ldots, \(k+2019\) are all divisible by squares. It will suffice to show there are infinitely many \(k\) in the arithmetic sequence \(i\mapsto P^2i+c\) such that \(k\) and \(k+2020\) are both squarefree.

Observe that for fixed primes \(q\), the density of positive integers \(i\) for which \(q^2\mid P^2i+c\) is \(\le1/q^2\). Similarly \(i\) with \(q^2\mid P^2i+c+2020\) have density \(\le1/q^2\), so the density of \(i\) for which \(q^2\) divides either \(P^2i+c\) or \(P^2i+c+2020\) is \(\le2/q^2\). But the density of \(i\) for which there exists \(q\) that divide either \(P^2i+c\) or \(P^2i+c+2020\) is at most \[\sum_q\frac2{q^2}<1,\]so there are infinitely many \(i\) for which \(P^2i+c\) and \(P^2i+c+2020\) are squarefree, as desired.

Basically the same as the China TST 2015 problem, except that you have to find a type of numbers that work, which is easy to do with CRT.

We will show that there are infinitely many positive integers $m$ such that $m$ and $m+2020$ are squarefree but $m+1,\ldots,m+2019$ are not. Pick massive distinct primes $p_1,\ldots,p_{2019}\gg 10^{987654321}$ and consider $m$ obeying the congruences \begin{align*} m &\equiv -1 \pmod{p_1^2}\\ m &\equiv -2 \pmod{p_2^2}\\ &\vdots\\ m &\equiv -2019 \pmod{p_{2019}^2}. \end{align*}Then $p_i^2$ doesn't divide $m$ or $m+2020$ for all $1 \leq i \leq 2019$. Suppose $m$ is of the form $Ax+B$. If $p \not \in \{p_1,\ldots,p_{2019}\}$, it is clear that the set of $x$ with $p^2 \mid Ax+B$ is $1/p^2$, and the same result is true for $Ax+B+2020$. Thus the density of $x$ such that there exists a prime $p$ with $p^2$ dividing either $Ax+B$ or $Ax+B+2020$is at most $$\sum_{p \text{ prime}} \frac{2}{p^2}<1,$$using a well-known result, so there are infinitely many $x$ with $Ax+B$ and $Ax+B+2020$ squarefree, as desired. $\blacksquare$

Wild. Consider the residue $a \pmod{p_1p_2\dots p_{2019}}$ such that $a \equiv -i \pmod{p_i}$ and $p_i \nmid 2020 - i$. Define $P = p_1p_2 \dots p_{2019}$. It remains to show that there are infinitely many pairs such that $a + kP, a + 2020 + kP$ are both squarefree. Consider the interval $\{1, 2, \dots, N\}$ for large $N$. Then there are at least \[ K = \left\lfloor \frac{N}{P} \right\rfloor - 3 \]such sequences that lie in the interval. Now, consider the $(1 + o(1)) \cdot \frac{N}{\log N} - 2019$ primes $p_{2020}, p_{2021}, \dots, p_k$. in the interval. Each prime $p_i$ can only appear at most $2 \cdot \left\lceil \frac{K}{p_i^2} \right\rceil$ times in any sequence, so in total they can only appear $2 \cdot \left(k + K \cdot P(2)\right)$ times where $P$ is the prime zeta function. There is as such at least guarenteed $(1 - P(2)) \cdot \left\lfloor \frac{N}{P} \right\rfloor + O\left(\frac{N}{\ln N}\right)$ which due to pigeonhole guarentees at least $(1 - P(2)) \cdot \left\lfloor \frac{N}{P} \right\rfloor$ pairs. Taking an arbitrarily large $N$ suffices.