Let out sequence is $a_1, a_2, ... , a_{40}, ...$ and notice that after $a_{40}$ we must have $a_1$ again, then $a_2$, etc. So, our sequence is repeated by the first $40$ elements. By the second condition we have that $a_i + a_{i+1} \neq 0$. Now we have $a_1=a_2=...=a_{20}=x$ and $a_{21}=a_{22}=...=a_{40}=-x$, where $x\in \{-1,1\}$. One can see that this type of sequence satisfies to the condition of the problem. Write $n=40m+k$, where $m\geqslant 1$ and $40>k\geqslant 0$. We don't care about sum of first $m$ $40$ tuples. So our problem asked us to find the maximum of the sum of first $k$ elements of the $a_1,a_2,...,a_{40}$. Of course, its happen when $k=20$ and $x=1$, so the answer is $\boxed{20}$