For each positive integer $n$ larger than $1$ with prime factorization $p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$, its [i]signature[/i] is defined as the sum $\alpha_1+\alpha_2+\cdots+\alpha_k$. Does there exist $2020$ consecutive positive integers such that among them, there are exactly $1812$ integers whose signatures are strictly smaller than $11$?
1. **Define the signature function and the problem setup:**
   Let $\sigma(n)$ be the signature of $n$, defined as the sum of the exponents in the prime factorization of $n$. We need to find if there exist $2020$ consecutive positive integers such that exactly $1812$ of them have signatures strictly smaller than $11$.

2. **Define the function $f(n)$:**
   Let $f(n) = |\{k \in [n, n+2019] \mid \sigma(k) < 11\}|$. This function counts the number of integers in the interval $[n, n+2019]$ whose signatures are less than $11$.

3. **Initial value of $f(n)$:**
   Notice that for small values of $n$, specifically $n \leq 2020$, the signature $\sigma(k)$ for $k \leq 2020$ is at most $\log_2(k)$. Since $\log_2(2020) < \log_2(2048) = 11$, we have $\sigma(k) < 11$ for all $k \leq 2020$. Therefore, $f(1) = 2020$.

4. **Constructing an $n$ such that $f(n) = 0$:**
   We can find an $n$ such that $f(n) = 0$ using the Chinese Remainder Theorem (CRT). Choose $2020$ distinct primes $p_1, p_2, \ldots, p_{2020}$ and construct $n$ such that:
   \[
   n \equiv 0 \pmod{p_1^{11}}, \quad n+1 \equiv 0 \pmod{p_2^{11}}, \quad \ldots, \quad n+2019 \equiv 0 \pmod{p_{2020}^{11}}
   \]
   This ensures that $\sigma(n+k) \geq 11$ for every $k = 0, 1, \ldots, 2019$, thus $f(n) = 0$.

5. **Discrete continuity of $f(n)$:**
   The function $f(n)$ can change by at most $1$ when $n$ is incremented by $1$. This is because the interval $[n, n+2019]$ shifts by one position, potentially adding one new integer with $\sigma < 11$ and removing one old integer with $\sigma < 11$.

6. **Intermediate value theorem for discrete functions:**
   Since $f(1) = 2020$ and there exists an $n$ such that $f(n) = 0$, and $f(n)$ changes by at most $1$ for each increment of $n$, by the intermediate value theorem for discrete functions, there must exist some $n$ such that $f(n) = 1812$.

Conclusion:
\[
\boxed{\text{True}}
\]