Hello IMOment 2

If $a,b,c$ form an arithmetic progression of integers, then $a$ and $c$ have the same parity. Moreover, any two numbers $a,c$ of the same parity can be the first and third elements of an arithmetic progression of integers. Thus there are $\binom{13}{2} + \binom{12}{2} = 144$ triplets of numbers between $1$ and $25$ that form an arithmetic progression.

For any even number $x$ , there are $11$ triplets that form an arithmetic progression where $x$ is either the first or the third element (just choose another even number to be the other terminal integer). There are also $\mathrm{min}(x-1,25-x)$ (the smaller of the numbers of integers less than or greater than $x$ ) triplets that form an arithmetic progression where $x$ is the second element.

Thus there are $144 - 11 - \mathrm{\min}(x,25-x) = 133 - \mathrm{min}(x,25-x)$ triplets that can be chosen that form an arithmetic progression and do not contain $x$ (for any even number $x$ ). Thus we obtain $122$ such triplets when $x = 10,16$ , making the answer $10 + 16 = \boxed{26}$ .

It is interesting to note that the solutions for any possible number of triplets (and not just $122$ ) are of the form $x,26-x$ and so add to $26$ .