= Solution
Work over the <finite field> $\mathbb F_p$. For each <set family> member form the <intersection polynomial>
$$
f_A(x_1,\ldots,x_n)=\prod_{i=1}^s\left(\sum_{a\in A}x_a-\ell_i\right).
$$
At the <characteristic vector of a set> $B$, its value is $\prod_i(|A\cap B|-\ell_i)$. The given restrictions make $f_A(\chi_B)=0$ for $A\neq B$ and $f_A(\chi_A)\neq0$. Hence these functions are <linearly independent>: evaluate a purported relation at each <set family> member to eliminate its coefficient.
Use <multilinear reduction on the Boolean cube>, replacing every positive power of $x_i$ by $x_i$. This preserves all evaluations on $\{0,1\}^n$ and cannot increase degree. The <multilinear polynomials> of degree at most $s$ have basis $x_S=\prod_{i\in S}x_i$ for $|S|\leq s$. Their dimension is $\sum_{j=0}^{\min(s,n)}\binom nj$, and therefore
$$
\boxed{|\mathcal A|\leq\sum_{j=0}^s\binom nj},
$$
where $\binom nj=0$ for $j>n$. The proof also covers a list whose distinct integers have repeated residues: no assumption of distinct residues is needed for this first bound.
To obtain the sharper conclusion, suppose the common size residue $r$ avoids $0,\ldots,s-1$. If $s>n$, no set on this ground set can have the required residue, so the <set family> is empty. Otherwise, add the <modular-size auxiliary polynomials>
$$
q_S(x)=x_S\left(\sum_{i=1}^n x_i-r\right),\qquad |S|\leq s-1,
$$
again understood as functions after <multilinear reduction on the Boolean cube>. They have degree at most $s$, and all vanish on the <set family>'s <characteristic vectors of sets>.
The $q_S$ are <linearly independent>. Indeed, evaluate $\sum_S c_Sq_S=0$ on $\chi_T$ with $|T|\leq s-1$:
$$
(|T|-r)\sum_{S\subseteq T}c_S=0.
$$
The first factor is nonzero in the <finite field>. Starting with $T=\varnothing$ and proceeding in increasing <cardinality> shows $c_T=0$ for every such $T$.
Moreover, the combined list consisting of the $f_A$ and the $q_S$ is <linearly independent>. Evaluation at each $\chi_A$ first kills the coefficient of $f_A$; the preceding argument then kills all remaining coefficients. Comparing this list with the dimension of the degree-at-most-$s$ space yields
$$
|\mathcal A|+\sum_{j=0}^{s-1}\binom nj\leq\sum_{j=0}^s\binom nj,
\qquad \boxed{|\mathcal A|\leq\binom ns}.
$$
This is a nonuniform common-residue version of the <Frankl-Wilson theorem>; exact equality of all member sizes was never used.
\b[The exclusion of the small residues is necessary.] A concrete example with $s<n$ is $n=4$, $p=5$, $s=3$, the residue list $0,1,3$, and the <set family> of all two-element subsets. Every member has size residue $r=2$, outside that list, while distinct members intersect in zero or one element. But
$$
|\mathcal A|=\binom42=6>\binom43=4.
$$
Here the omitted condition fails at $j=2$. Thus the example satisfies all the other hypotheses, rather than relying on a vacuous $s>n$ case.
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