Cute. The answer is $2n+1$. $1,2n+1,2,2n,3,2n-1,\ldots,n+1$ clearly works, just "jump" between the alternating sequences depending on whether you need a $+1$ or a $-1$. Now consider a $s_i$ with class $a_i$ that works for some $n$. We must have that it works for $1,1,\ldots,1$ and $-1,-1\ldots,-1$. Let the subsequence for the former be $t_1,\ldots,t_{n+1}$. Between $t_{i}<t_{i+1}$ in the sequence $s$, there must be at least one increase, and all of these intervals are mutually disjoint, so there must be $n$ $1$s in $a_i$. Similarly, there must be $n$ $-1s$. The conclusion is immediate.

Sorry if I made any mistake, but can't we just consider the sequences $a_1<a_2<...<a_{2n}$ or $a_1>a_2>...>a_{2n}$ (which gives $m=2n$) and then take the terms depend on you want $1$ or $-1$, hence QED.