Solution to Exercise 6 Let $d_{i}=a_{i+1}-a_{i}$. Let $m$ be such that $a_{m}$ is the maximum. Then we have $d_{1}>d_{2}>\cdots>d_{m-1}>0>d_{m}>\cdots>d_{124}$.

Suppose that $d_{m-1}=1$ and $d_{m}=-1$. Then $a_{m-1}=a_{m+1}$, which contradicts the fact that the 125 integers are all distinct. Therefore, either $d_{m-1} \geqslant 2$ or $d_{m} \leqslant-2$. Without loss of generality, we can assume that $d_{m-1} \geqslant 2$. Then we have $a_{m}=a_{1}+\left(d_{1}+\cdots+d_{m-1}\right) \geqslant 1+\cdots+m$.

Similarly, $a_{m}=a_{125}-\left(d_{m}+\cdots+d_{124}\right) \geqslant 1+(1+2+\cdots+(125-m))$.

One of the numbers $m$ or $125-m$ is $\geqslant 63$, so $a_{m} \geqslant 1+\cdots+63=2016$.

This number can be achieved by taking $a_{1}=1$ and $d_{i}=64-i$ for $i \leqslant 62$ and $d_{i}=62-i$ otherwise. Let's verify that this sequence of numbers works:

We have $1=a_{1}a_{64}>\cdots>a_{125}=63$. Furthermore, if there exists $i63$ then $0=d_{i}+\cdots+d_{62}+d_{63}+\cdots+d_{j-1}=2+\cdots+(64-i)-(1+\cdots+(j-63))$. Necessarily, $64-i>j-63$, and thus by simplifying we get $(j-63+1)+\cdots+(64-i)=1$, which is impossible. Therefore, the 125 integers are indeed distinct.
Exercise 6. Determine the smallest integer $N$ such that one can find 125 distinct integers $a_{1}, a_{2}, \ldots, a_{125}$ in a line so that:

(1) each of these integers is strictly positive, and less than or equal to $N$;

(2) each of the 123 integers $a_{2}, \ldots, a_{124}$ is strictly greater than the arithmetic mean of the integer written to its left and the integer written to its right.

N.B. The arithmetic mean of $a$ and $b$ is the number $\frac{a+b}{2}$.