For positive integer $k,$ we say that it is a [i]Taurus integer[/i] if  we can delete one element from the set $M_k=\{1,2,\cdots,k\},$ such that the sum of remaining $k-1$ elements is a positive perfect square. For example, $7$ is a Taurus integer, because if we delete $3$ from $M_7=\{1,2,3,4,5,6,7\},$ the sum of remaining $6$ elements is $25,$ which is a positive perfect square.
$(1)$ Determine whether $2021$ is a Taurus integer.
$(2)$ For positive integer $n,$ determine the number of Taurus integers in $\{1,2,\cdots,n\}.$
1. **Determine whether 2021 is a Taurus integer:**

   To determine if 2021 is a Taurus integer, we need to check if we can delete one element from the set \( M_{2021} = \{1, 2, \ldots, 2021\} \) such that the sum of the remaining 2020 elements is a perfect square.

   The sum of the first \( k \) positive integers is given by:
   \[
   S_k = \frac{k(k+1)}{2}
   \]
   For \( k = 2021 \):
   \[
   S_{2021} = \frac{2021 \times 2022}{2} = 2021 \times 1011 = 2043231
   \]

   We need to check if there exists an integer \( x \) in \( M_{2021} \) such that:
   \[
   S_{2021} - x = n^2 \quad \text{for some integer } n
   \]
   This implies:
   \[
   2043231 - x = n^2
   \]
   Rearranging, we get:
   \[
   x = 2043231 - n^2
   \]
   We need \( x \) to be an integer in the range \( 1 \leq x \leq 2021 \). Therefore:
   \[
   1 \leq 2043231 - n^2 \leq 2021
   \]
   This simplifies to:
   \[
   2041210 \leq n^2 \leq 2043230
   \]
   Taking the square root of the bounds:
   \[
   \sqrt{2041210} \leq n \leq \sqrt{2043230}
   \]
   Approximating the square roots:
   \[
   1428.6 \leq n \leq 1429.5
   \]
   Since \( n \) must be an integer, we check \( n = 1429 \):
   \[
   n^2 = 1429^2 = 2042041
   \]
   Then:
   \[
   x = 2043231 - 2042041 = 1190
   \]
   Since \( 1190 \) is within the range \( 1 \leq x \leq 2021 \), 2021 is a Taurus integer.

2. **Determine the number of Taurus integers in \( \{1, 2, \ldots, n\} \):**

   We need to determine for which \( k \) in \( \{1, 2, \ldots, n\} \), there exists an element \( x \) in \( M_k \) such that the sum of the remaining \( k-1 \) elements is a perfect square.

   The sum of the first \( k \) positive integers is:
   \[
   S_k = \frac{k(k+1)}{2}
   \]
   We need to find \( x \) such that:
   \[
   S_k - x = n^2 \quad \text{for some integer } n
   \]
   This implies:
   \[
   x = S_k - n^2
   \]
   We need \( x \) to be an integer in the range \( 1 \leq x \leq k \). Therefore:
   \[
   1 \leq \frac{k(k+1)}{2} - n^2 \leq k
   \]
   This simplifies to:
   \[
   \frac{k(k+1)}{2} - k \leq n^2 \leq \frac{k(k+1)}{2} - 1
   \]
   Simplifying further:
   \[
   \frac{k(k-1)}{2} \leq n^2 \leq \frac{k(k+1)}{2} - 1
   \]
   For large \( k \), the range of possible \( n^2 \) values is large enough to include at least one perfect square. However, for small \( k \), we need to check manually.

   By manual checking, we find that \( k = 5 \) and \( k = 8 \) are not Taurus integers. For all other \( k \), there exists an \( x \) such that \( S_k - x \) is a perfect square.

   Therefore, the number of Taurus integers in \( \{1, 2, \ldots, n\} \) is:
   \[
   n - 2 \quad \text{(excluding 5 and 8)}
   \]

The final answer is \( \boxed{ n - 2 } \)