Yes. Let $f(n)$ be the number of integers among $n+1, n+2, \dots, n+2020$ having signatures less than $11$. Since $2^{11} = 2048 > 2021$, all of $2, 3, \dots, 2021$ have signatures smaller than $11$. Therefore, we have $f(1) = 2020$.

Next, let $p_1, p_2, \dots, p_{2020}$ be distinct primes. By the Chinese remainder theorem,

there exists a positive integer $m$ such that
$$\text{l@ l@ l m -1 p 1 11 , m -2 p 2 11 , m -2020 p 2020 11 .}$$
For this $m$, since $p_j^{11}$ divides $m+j$ for $j = 1, 2, \dots, 2020$, each of the integers $m+1, m+2, \dots, m+2020$ has signature larger than or equal to $11$. This implies $f(m) = 0$.

Now, when we move from the integers $n+1, n+2, \dots, n+2020$ to the integers $n+2, n+3, \dots, n+2021$, the number of integers with signature less than $11$ will be increased or decreased by $1$, or remain unchanged. Thus, the difference between $f(n)$ and $f(n+1)$ is at most $1$. Therefore, in order to decrease $f(1) = 2020$ to $f(m) = 0$, we must come across a positive integer $k$ such that $f(k) = 1812$, and we are done.