A [i]squiggle[/i] is composed of six equilateral triangles with side length $1$ as shown in the figure below. Determine all possible integers $n$ such that an equilateral triangle with side length $n$ can be fully covered with [i]squiggle[/i]s (rotations and reflections of [i]squiggle[/i]s are allowed, overlappings are not).

[asy]
import graph; size(100); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; 
draw((0,0)--(0.5,1),linewidth(2pt)); draw((0.5,1)--(1,0),linewidth(2pt)); draw((0,0)--(3,0),linewidth(2pt)); draw((1.5,1)--(2,0),linewidth(2pt)); draw((2,0)--(2.5,1),linewidth(2pt)); draw((0.5,1)--(2.5,1),linewidth(2pt)); draw((1,0)--(2,2),linewidth(2pt)); draw((2,2)--(3,0),linewidth(2pt));
dot((0,0),ds); dot((1,0),ds); dot((0.5,1),ds); dot((2,0),ds); dot((1.5,1),ds); dot((3,0),ds); dot((2.5,1),ds); dot((2,2),ds); clip((-4.28,-10.96)--(-4.28,6.28)--(16.2,6.28)--(16.2,-10.96)--cycle);[/asy]
1. **Determine the area of the squiggle and the equilateral triangle:**
   - Each equilateral triangle with side length 1 has an area of:
     \[
     A_{\text{triangle}} = \frac{\sqrt{3}}{4} \times 1^2 = \frac{\sqrt{3}}{4}
     \]
   - A squiggle is composed of 6 such triangles, so its area is:
     \[
     A_{\text{squiggle}} = 6 \times \frac{\sqrt{3}}{4} = \frac{3\sqrt{3}}{2}
     \]
   - An equilateral triangle with side length \( n \) has an area of:
     \[
     A_{\text{big triangle}} = \frac{\sqrt{3}}{4} \times n^2
     \]

2. **Determine the condition for covering the big triangle with squiggles:**
   - The area of the big triangle must be an integer multiple of the area of a squiggle:
     \[
     \frac{\sqrt{3}}{4} \times n^2 = k \times \frac{3\sqrt{3}}{2}
     \]
     where \( k \) is an integer.
   - Simplifying, we get:
     \[
     \frac{n^2}{4} = k \times \frac{3}{2}
     \]
     \[
     n^2 = 6k
     \]
   - Therefore, \( n^2 \) must be a multiple of 6, implying:
     \[
     n \equiv 0 \pmod{6}
     \]

3. **Coloring argument for further restriction:**
   - Color the cells of the big triangle in a checkerboard pattern (alternating black and white).
   - Each squiggle covers 4 cells of one color and 2 cells of the other color.
   - The total number of cells in the big triangle is:
     \[
     \text{Total cells} = \frac{n(n+1)}{2} + \frac{n(n-1)}{2} = n^2
     \]
   - For the big triangle to be fully covered by squiggles, the number of cells of each color must be even.
   - This implies both \( \frac{n(n+1)}{2} \) and \( \frac{n(n-1)}{2} \) must be even, which means \( n \equiv 0 \pmod{4} \).

4. **Combining the conditions:**
   - From the area condition, \( n \equiv 0 \pmod{6} \).
   - From the coloring condition, \( n \equiv 0 \pmod{4} \).
   - The least common multiple of 6 and 4 is 12, so:
     \[
     n \equiv 0 \pmod{12}
     \]

5. **Verification for \( n = 12 \):**
   - For \( n = 12 \), the area of the big triangle is:
     \[
     A_{\text{big triangle}} = \frac{\sqrt{3}}{4} \times 12^2 = 36\sqrt{3}
     \]
   - The number of squiggles needed is:
     \[
     \frac{36\sqrt{3}}{\frac{3\sqrt{3}}{2}} = 24
     \]
   - Since 24 is an integer, it is possible to cover the big triangle with squiggles.

The final answer is \( \boxed{ n \equiv 0 \pmod{12} } \)