The hall in a castle is a regular hexagon where its sides' length is 6 meters. The floor of the hall is to be tiled with equilateral triangular tiles where its sides' length is 50 centimeters. Each tile is divided into three congruent triangles by their altitudes up to its orthocenter (see below). Each of these small triangles are colored such that each tile has different colors and no two tiles have identical colorings. How many colors at least are required?

A tile's pattern is:
[asy]
draw((0,0.000)--(2,0.000));
draw((2,0.000)--(1,1.732));
draw((1,1.732)--(0,0.000));
draw((1,0.577)--(0,0.000));
draw((1,0.577)--(2,0.000));
draw((1,0.577)--(1,1.732));
[/asy]
1. **Calculate the number of tiles:**
   - The hall is a regular hexagon with side length 6 meters.
   - Each tile is an equilateral triangle with side length 50 centimeters (0.5 meters).
   - The area of the hexagon can be calculated using the formula for the area of a regular hexagon: 
     \[
     A_{\text{hexagon}} = \frac{3\sqrt{3}}{2} s^2
     \]
     where \( s \) is the side length of the hexagon.
     \[
     A_{\text{hexagon}} = \frac{3\sqrt{3}}{2} \times 6^2 = 54\sqrt{3} \text{ square meters}
     \]

2. **Calculate the area of one tile:**
   - The area of an equilateral triangle with side length \( a \) is given by:
     \[
     A_{\text{triangle}} = \frac{\sqrt{3}}{4} a^2
     \]
     where \( a = 0.5 \) meters.
     \[
     A_{\text{triangle}} = \frac{\sqrt{3}}{4} \times (0.5)^2 = \frac{\sqrt{3}}{16} \text{ square meters}
     \]

3. **Calculate the number of tiles needed:**
   - The number of tiles is the total area of the hexagon divided by the area of one tile:
     \[
     \text{Number of tiles} = \frac{A_{\text{hexagon}}}{A_{\text{triangle}}} = \frac{54\sqrt{3}}{\frac{\sqrt{3}}{16}} = 54 \times 16 = 864
     \]

4. **Determine the number of colors required:**
   - Each tile is divided into three congruent triangles, and each of these small triangles must be colored such that no two tiles have identical colorings.
   - We need to find the minimum number of colors \( n \) such that the number of ways to color the tiles is at least 864.
   - The number of ways to color each tile with \( n \) colors is given by the combination:
     \[
     \binom{n}{3} = \frac{n!}{3!(n-3)!}
     \]
   - We need to solve the inequality:
     \[
     \binom{n}{3} \geq 864
     \]
   - Testing values of \( n \):
     \[
     \binom{18}{3} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 816 \quad (\text{not sufficient})
     \]
     \[
     \binom{19}{3} = \frac{19 \times 18 \times 17}{3 \times 2 \times 1} = 969 \quad (\text{sufficient})
     \]
   - Therefore, \( n \geq 19 \).

The final answer is \( \boxed{ n \geq 19 } \).