National Olympiad Final Round

The equation $|a-b|=|b-c|$ is satisfied if and only if either $a-b=b-c$ or $a-b=c-b$. The first equality is equivalent to $a+c=2b$, the second is equivalent to $a=c$.

To fulfill the condition $a+c=2b$, the numbers $a$ and $c$ must have the same parity. Then their sum is even and $b$ lies between $a$ and $c$, whence it also falls between the required bounds. There are $19$ possibilities for choosing $a$ or $c$ as an even number ($0,2,\dots ,36$ are suitable) which gives $361$ possibilities in total. The number of possibilities to choose an odd number $a$ or $c$ is $18$ ($1,3,\dots ,35$ are suitable) which gives $324$ possibilities in total. There are therefore $685$ triples satisfying $a+c=2b$.

The number of triples satisfying $a=c$ is $1369$, since $a$ and $b$ can be chosen arbitrarily. There are $37$ solutions that satisfy both $a+c=2b$ and $a=c$ and hence being counted twice, since these two conditions hold simultaneously if and only if $a=b=c$.

Consequently, the total number of solutions meeting the conditions of the problem is $685+1369-37$ which equals $2017$.