, 2022

Solution:
We claim the maximum is $5$, achieved by the sequence $(1,3,5,7,2,4,6,8)$. Now we prove that we cannot do better.

Since $a_{1}+a_{2}+\ldots+a_{8}=1+2+\ldots+8=36$, then there are at most $6$ squares in
$$\left\{a_{1}, a_{1}+a_{2}, \ldots, a_{1}+a_{2}+\cdots+a_{8}\right\}.$$
Note that if $a_{1}+\cdots+a_{k}=n^{2}$ and $a_{1}+\cdots+a_{j}=(n+1)^{2}$, then $a_{k+1}+\ldots+a_{j}=2n+1$. Since $2n+1$ is odd, $a_{m}$ must be odd for some $m \in [k+1, j]$.

Thus, if all of $1,4,9,16,25$, and $36$ are in
$$\left\{a_{1}, a_{1}+a_{2}, \ldots, a_{1}+a_{2}+\cdots+a_{8}\right\}$$
then $a_{1}=1$, and there are five more odd values in $\left\{a_{1}, a_{2}, \ldots, a_{8}\right\}$, which is a contradiction because there are only four odd numbers in $\{1,2, \ldots, 8\}$.