this question definitely had our team involved in a mathematical debate. nobody knew who was right. Alaban, our teammate told us there were no roots. Some answered 4, that's me and Jako... so need help here...

For (1), draw the graph of $|x + 1| + |x + 2| + |x + 3| = y$, we find that there can be 0,1 or 2 roots. For (2), I think case division should work (eg. the difference $d = 0$, $1$ or more than $2$)

For (2), $n=26$, right?

enescu, you are right!

Can someone post the solution to 2?

First, observe that if $a_{1}<a_{2}<\ldots<a_{n}$ are real numbers and $f:\mathbb{R}\rightarrow\mathbb{R}$ is defined as \[ f\left( x\right) =\left| x-a_{1}\right| +\left| x-a_{2}\right| +\ldots+\left| x-a_{n}\right| , \] then $f$ is first decreasing, and then increasing. More precisely, if $n$ is odd, say $n=2k+1,$ then $f$ is strictly decreasing on the interval $(-\infty,a_{k}]$ and strictly increasing on $[a_{k},+\infty)$, while if $n$ is even, say $n=2k,$ then $f$ is strictly decreasing on the interval $(-\infty,a_{k}],$ constant on $[a_{k},a_{k+1}]$ and strictly increasing on $[a_{k+1},+\infty).$ Now, let WLOG $a_{1}\leq\ldots\leq a_{n}$ be our arithmetic sequence and $f$ defined as above. We see that $f(0)=f\left( -1\right) =f\left( 2\right)$, hence three values of $f$ are equal. This is possible only if $n$ is even (say $n=2k$) and $0,-1,2\in\lbrack a_{k},a_{k+1}].$ We deduce that the common difference $d$ is at least $3,$ and $a_{k}\leq-1,a_{k+1}\geq2.$ A short computation shows that in this case \[ 507=\sum\left| a_{i}\right| \geq3k^{2}, \] hence $k\leq13,$ that is, $n\leq26.$ Equality is obtained when $a_{1}=-37,a_{2}=-34,\ldots,a_{26}=38.$