Given a set $A=\{1;2;...;4044\}$. They color $2022$ numbers of them by white and the rest of them by black. With each $i\in A$, called the [b][i]important number[/i][/b] of $i$ be the number of all white numbers smaller than $i$ and black numbers larger than $i$. With every natural number $m$, find all positive integers $k$ that exist a way to color the numbers that can get $k$ important numbers equal to $m$.
1. **Define the problem and notation:**
   Given a set \( A = \{1, 2, \ldots, 4044\} \), we color 2022 numbers white and the rest black. For each \( i \in A \), the important number of \( i \) is defined as the number of white numbers smaller than \( i \) plus the number of black numbers larger than \( i \). We need to find all positive integers \( k \) such that there exists a way to color the numbers to get \( k \) important numbers equal to \( m \).

2. **Set up the problem:**
   Let \( N = 2022 \). For each \( 0 \le k \le 2N-1 \), the number of numbers with weight \( k \) can be an even number from \( 0 \) to \( 2\min\{k+1, 2N-k\} \).

3. **Define the coloring function:**
   Let \( c_i = 1 \) if the number \( i \) is colored black, and \( c_i = -1 \) otherwise. This means \( c_1 + c_2 + \cdots + c_{2N} = 0 \).

4. **Calculate the number of white and black numbers:**
   The number of white numbers among the first \( t-1 \) is:
   \[
   \frac{(t-1) - (c_1 + c_2 + \cdots + c_{t-1})}{2}
   \]
   The number of black numbers among the last \( 2N-t \) is:
   \[
   \frac{c_{t+1} + c_{t+2} + \cdots + c_{2N} + 2N - t}{2}
   \]

5. **Sum of white and black numbers:**
   Their sum is also the weight of \( t \). This is equal to:
   \[
   f(t) = \frac{2N-1}{2} - \frac{1}{2}(c_1 + c_2 + \cdots + c_{t-1} + c_1 + c_2 + \cdots + c_t) = \frac{2N-1 - c_t}{2} - (c_1 + c_2 + \cdots + c_{t-1})
   \]

6. **Claim:**
   For a fixed \( k \), the number of solutions to \( f(t) = k \) is even. When the solutions are written in ascending order, the colors alternate between black and white.

7. **Proof of the claim:**
   Let \( d_j = \frac{c_j + c_{j-1}}{2} = f(j-1) - f(j) \). Suppose \( f(t) = f(u) = k \) and \( c_t = c_u = 1 \). Then there exists \( v \) such that \( t < v < u \), \( c_v = -1 \), and \( f(v) = k \). If \( c_{u-1} = -1 \), then \( u-1 \) satisfies the criteria. Otherwise, \( c_{u-1} = c_{t+1} = 1 \), \( f(t+1) = k-1 \), and \( f(u-1) = k+1 \). Since \( d_j \in \{-1, 0, 1\} \), \( f \) is discrete continuous. Thus, there exists \( v \) such that \( f(v) > f(v-1) \), \( f(v) = k \), which implies \( d_v = -1 \), \( c_v = c_{v-1} = -1 \).

8. **Leftmost and rightmost solutions:**
   The leftmost and rightmost solutions to \( f(t) = k \) have different colors. Observe that \( f(1) = \frac{2N-1 - c_1}{2} \) and \( f(n) = \frac{2N-1 + c_n}{2} \). Therefore, by discrete continuity, if \( k \le N-1 \), the smallest solution to \( f(t) = k \) is black and the largest number is white. If \( k \ge N \), the smallest solution is white while the largest is black.

9. **Constructing solutions:**
   For \( 0 \le k \le N-1 \), there are at most \( k+1 \) solutions to \( f(t) = k \) where \( t \) is white. If we sort the white numbers as \( w_1 < w_2 < \cdots < w_N \), then only \( w_1, w_2, \ldots, w_{k+1} \) can qualify, since all other numbers have at least \( k+1 \) white numbers to their left. Similarly, there are at most \( k+1 \) solutions to \( f(t) = k \) where \( t \) is black.

10. **Constructing for weight \( 0 \le k \le N-1 \) with \( 0 \le \frac{m}{2} \le k+1 \) solutions:**
    For \( m = 0 \), use the construction \( c_j = (-1)^j \), which ensures all numbers have weight \( N \).

    Otherwise, set \( c_1 = c_2 = \cdots = c_{N-(k+1)} = 1 \), then \( c_{N-k-1+t} = (-1)^{t+1} \) for \( 1 \le t \le m \). Then color the next \( N - \frac{m}{2} \) numbers white, and color the final remaining numbers black. This construction works.

The final answer is \( \boxed{ k } \) such that \( 0 \le k \le 2N-1 \).