An equilateral triangle $T$ with side length $2022$ is divided into equilateral unit triangles with lines parallel to its sides to obtain a triangular grid. The grid is covered with figures shown on the image below, which consist of $4$ equilateral unit triangles and can be rotated by any angle $k \cdot 60^{\circ}$ for $k \in \left \{1,2,3,4,5 \right \}$.
The covering satisfies the following conditions:

$1)$ It is possible not to use figures of some type and it is possible to use several figures of the same type. The unit triangles in the figures correspond to the unit triangles in the grid.
$2)$ Every unit triangle in the grid is covered, no two figures overlap and every figure is fully contained in $T$.

Determine the smallest possible number of figures of type $1$ that can be used in such a covering.

[i]Proposed by Ilija Jovcheski[/i]
To solve this problem, we need to determine the smallest possible number of figures of type 1 that can be used to cover an equilateral triangle \( T \) with side length 2022, divided into unit equilateral triangles.

1. **Calculate the total number of unit triangles in \( T \)**:
   The number of unit triangles in an equilateral triangle with side length \( n \) is given by the formula:
   \[
   \text{Total unit triangles} = \frac{n(n+1)}{2}
   \]
   For \( n = 2022 \):
   \[
   \text{Total unit triangles} = \frac{2022 \times 2023}{2} = 2022 \times 1011 = 2046246
   \]

2. **Determine the excess of upward-pointing unit triangles**:
   In an equilateral triangle, half of the unit triangles are upward-pointing and half are downward-pointing, except for the extra row at the bottom. Therefore, the number of upward-pointing unit triangles is:
   \[
   \text{Upward-pointing triangles} = \frac{2046246 + 2022}{2} = 1024134
   \]
   The number of downward-pointing unit triangles is:
   \[
   \text{Downward-pointing triangles} = \frac{2046246 - 2022}{2} = 1022112
   \]
   The excess of upward-pointing unit triangles is:
   \[
   1024134 - 1022112 = 2022
   \]

3. **Analyze the figures**:
   - Figures of type 1 have an excess of \( \pm 2 \) upward-pointing unit triangles depending on their orientation.
   - Figures of types 2, 3, and 4 have an equal number of upward- and downward-pointing unit triangles.

4. **Determine the minimum number of figures of type 1 needed**:
   Since figures of types 2, 3, and 4 do not affect the balance of upward- and downward-pointing unit triangles, the excess of 2022 upward-pointing unit triangles must be resolved by figures of type 1. Each figure of type 1 can resolve an excess of \( \pm 2 \) upward-pointing unit triangles. Therefore, the minimum number of figures of type 1 required is:
   \[
   \frac{2022}{2} = 1011
   \]

5. **Construct a covering using induction**:
   - For a triangle of side length \( 2n \), decompose it into a smaller triangle of side length \( 2(n-1) \) and an isosceles trapezoid.
   - Cover the trapezoid with one figure of type 1 and \( 2n-2 \) figures of type 2 rotated \( 60^\circ \) clockwise.
   - Inductively, this shows that a triangle of side length \( 2n \) can be covered using \( n \) figures of type 1.

Putting all these steps together, we conclude that the smallest number of figures of type 1 that can be used in any legitimate covering of \( T \) is \( 1011 \).

The final answer is \(\boxed{1011}\)