Maximum Average of Six Natural Numbers

Let the six distinct natural numbers in increasing order be $a_1 < a_2 < a_3 < a_4 < a_5 < a_6$.

• Average of 2 smallest is 14 $\implies a_1 + a_2 = 28$.
• Average of 2 largest is 28 $\implies a_5 + a_6 = 56$.

To maximize the overall average, we must maximize $a_3 + a_4$. Since all numbers are distinct natural numbers: To maximize $a_3$ and $a_4$, we must make $a_5$ as large as possible, which means $a_6$ is as small as possible. Since $a_4 < a_5 < a_6$ and $a_5 + a_6 = 56$: If $a_5 = 27$, then $a_6 = 29$. (Since $a_5 < a_6$ and integers). Then $a_4 \le 26$ and $a_3 \le 25$.

Sum of all six numbers $= (a_1 + a_2) + a_3 + a_4 + (a_5 + a_6) = 28 + a_3 + a_4 + 56 = 84 + a_3 + a_4$. Max possible sum $= 84 + 25 + 26 = 135$.

Max possible average $= \frac{135}{6} = 22.5$.