Determine the smallest positive integer $k{}$ satisfying the following condition: For any configuration of chess queens on a $100 \times 100$ chequered board, the queens can be coloured one of $k$ colours so that no two queens of the same colour attack each other.

[i]Russia, Sergei Avgustinovich and Dmitry Khramtsov[/i]
1. **Base Case:**
   - For the base case, consider a configuration with at most 5 queens. Since no two queens can attack each other, we can assign each queen a unique color. Thus, 5 colors suffice for this case.

2. **Induction Hypothesis:**
   - Assume that for any configuration of \( n \) queens on a \( 100 \times 100 \) board, 5 colors are sufficient such that no two queens of the same color attack each other.

3. **Induction Step:**
   - Consider a configuration with \( n+1 \) queens. Remove the leftmost queen \( Q \) among the topmost ones. By the induction hypothesis, the remaining \( n \) queens can be colored with 5 colors such that no two queens of the same color attack each other.
   - When we return \( Q \) to the board, it can attack at most 4 other queens. Since there are 5 colors available, we can assign \( Q \) a color that is different from the colors of the queens it attacks. Thus, 5 colors suffice for \( n+1 \) queens.

4. **Proof by Contradiction for \( k < 5 \):**
   - To show that 4 colors are not sufficient, consider the arrangement of queens on the shaded squares as shown in the problem statement. Assume that the queens can be colored with 4 colors such that no two queens of the same color attack each other.
   - **Lemma:** Either the queens \( a_1, a_2, a_3, a_4, a_5 \) or the queens \( a_1, a_2', a_3', a_4', a_5' \) are assigned two colors alternately.
     - **Proof of Lemma:** Assume the colors of \( a_1, \ldots, a_5 \) do not follow the pattern. Then three consecutive queens on this diagonal have three distinct colors, say, 1, 2, and 3. Any four queens forming a diamond should have four distinct colors. This means that the north-east diagonals starting at the three queens should be colored as shown in the middle figure. In this coloring, each north-east diagonal contains only two colors, while each north-west diagonal contains three.
     - Similarly, if the colors of \( a_1, a_2', a_3', a_4', a_5' \) do not follow the pattern, then the north-west diagonals starting at three consecutive queens on this diagonal are colored in a similar manner, but with reversed roles of the diagonals. So the two obtained colorings do not agree where they meet.
   - Applying the lemma and using the symmetry of the arrangement, assume that the queens \( a_1, \ldots, a_5 \) are assigned colors 1 and 2 alternately, so that \( a_1 \) and \( a_5 \) share color 1. Then the queens \( b \) and \( c \) should have colors 3 and 4 (in some order), and hence the queens \( d \) and \( e \) should both have color 2. This is impossible since \( d \) and \( e \) attack each other.

5. **Conclusion:**
   - The problem calls for a proper coloring of a graph with queens as vertices, where two queens share an edge if they attack each other. The argument in the first paragraph is an instance of a well-known fact stating that any \( k \)-degenerate graph is properly \( (k + 1) \)-colorable.
   - For the remaining part of the solution, there are many different arrangements showing that \( k \geq 5 \). One of them is depicted in the right figure above. To show that it works, notice that either \( a, b, c \) or \( a, b', c' \) have three distinct colors (otherwise \( c \) and \( c' \) share the color of \( a \)). Assuming the former, we get, as in the above lemma, that \( a \) and the queens to the right of it are assigned two colors alternately, say 1 and 2. Then the queens \( d, e \), and \( f \) should all have colors 3 or 4, so two of them share the same color. This is impossible.

\(\blacksquare\)

The final answer is \( \boxed{ k = 5 } \)