= Solution
Again let $v_i\in\mathbb F_2^n$ be the characteristic vectors. They all lie in the even-weight subspace
$$
E=\{v\in\mathbb F_2^n:v\mathbin\cdot\mathbf1=0\},
$$
which has <dimension of a vector space>[dimension] $n-1$. Their <Gram matrix> is
$$
G=(v_i\mathbin\cdot v_j)_{i,j}=J_m+I_m,
$$
because its diagonal entries are zero and its off-diagonal entries are one. If $Gx=0$ and $s=\sum_i x_i$, then $x=s\mathbf1$. Consequently
$$
\operatorname{rank}G=
\begin{cases}
m,&m\text{ even},\\
m-1,&m\text{ odd}.
\end{cases}
$$
Since $G=VV^T$ for a matrix whose rows lie in $E$, its rank is at most $n-1$. If $n$ is even, an even $m$ cannot reach $n$, while an odd $m$ is at most $n-1$; hence $m\leq n-1$. If $n$ is odd, the same calculation gives $m\leq n$.
Both bounds are sharp. For odd $n$, take $A_i=[n]\setminus\{i\}$ for $1\leq i\leq n$: each set has even size and two distinct sets meet in the odd number $n-2$. For even $n$, apply the same construction to $[n-1]$ and regard the resulting $n-1$ sets as subsets of $[n]$.
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