, 2023

Number the examiners from $1$ to $10$ and consider the results of contestants as binary sequences of the form $x_1x_2\ldots x_{10}$, in which $x_i = 0$ or $1$ if the $i$th examiner gave pass or fail response for the corresponding candidate. We can directly construct $24$ binary strings that satisfy the “separative” condition. Indeed, choose strings of the form $(x_1x_2x_3x_4x_5x_6)(x_7x_8x_9x_{10})$, where: $x_1x_2x_3x_4x_5x_6$ contains exactly $3$ zeros and $3$ ones; and $x_7x_8x_9x_{10}$ will contain all binary strings of length $4$. Since $\binom{6}{3} = 20$ and two strings $x_1x_2x_3x_4x_5x_6 \neq x'_1x'_2x'_3x'_4x'_5x'_6$ would be different at at least $2$ positions, so the rest just need to differ by at least $1$ position. Since $2^4 = 16$, there will be $4$ repeated strings at the last $4$ positions, we just need to choose those strings that differ by at least $3$ positions out of the first $6$ positions. For detail,
$$\begin{array}{lcl} (000111)(0000) & \rightarrow & (a) \\ (001011)(0001) & \rightarrow & (b) \\ (001101)(0010) & \rightarrow & (c) \\ (001110)(0011) & \rightarrow & (d) \\ (010011)(0100) \\ (010101)(0101) \\ (010110)(0110) \\ (011001)(0111) \\ (011010)(1000) \\ (011100)(1001) \\ (100011)(1010) \\ (100101)(1011) \\ (100110)(1100) \\ (101001)(1101) \\ (101010)(1110) \\ (101100)(1111) \\ (110001)(0011) & \rightarrow & (d') \\ (110010)(0010) & \rightarrow & (c') \\ (110100)(0001) & \rightarrow & (b') \\ (111000)(0000) & \rightarrow & (a'). \end{array}$$
Finally, add $4$ more strings as the following:
$$\begin{array}{l} (000000)(0000), (000000)(1111), \\ (111111)(0000), (111111)(0000). \end{array}$$
It is easy to check that they are pairwise separative and also separative with the other $20$ strings. $\square$