= Solution
Assume first $1\le t\le n$, and write $d=n-t$. The sharp <nonuniform t-intersecting family bound> is
$$
\boxed{|\mathcal A|\le
\begin{cases}
\displaystyle\sum_{i=0}^q\binom ni,&d=2q,\\[3pt]
\displaystyle2\sum_{i=0}^q\binom{n-1}i,&d=2q+1.
\end{cases}}
$$
We prove the bound by deriving the needed <Kleitman diametric theorem>. Identify <subsets> with binary words, with <Hamming distance> $|A\triangle B|$. Let $D(n,d)$ be the two expressions above. A <set family> of <Hamming diameter> at most $d<n$ has size at most $D(n,d)$, as follows.
First apply <downward coordinate compression>: delete coordinate $i$ from a member only when its deletion is not already present. This preserves size and cannot increase the <Hamming diameter>. The only possible increased distance would be between a newly lowered member $A\setminus\{i\}$ and a retained member $B$ containing $i$. The deletion $B\setminus\{i\}$ must have belonged to the old <set family>, or $B$ would have been lowered too. The old pair $A,B\setminus\{i\}$ has precisely that increased distance, proving it was already allowed. Repeated downward compressions terminate in a <down-set>.
Next apply all <elementary set shifts> $S_{ij}$ with $i<j$. They preserve size and <Hamming diameter>. If shifting one member but retaining the other increased their distance, the retained member must contain $j$ but not $i$; its shifted partner was already present, and the old pair with that partner has the same distance. These shifts preserve the <down-set> property: a <subset> of a shifted member either was already present or is obtained by shifting its corresponding old <subset>. For a retained member, a <subset> cannot disappear, because the shifted partner of that <subset> is also a <subset> of the member or of its retained shifted partner. Repeated shifts terminate, since the sum of the selected coordinate labels decreases whenever a change occurs. We now have a shifted <down-set> $\mathcal F$ of the same size and <Hamming diameter>.
Split it into sections on $[n-1]$:
$$
\mathcal F_0=\{A:n\notin A,\ A\in\mathcal F\},\qquad
\mathcal F_1=\{A:A\cup\{n\}\in\mathcal F\}.
$$
The first section has <Hamming diameter> at most $d$. We claim the second has <Hamming diameter> at most $d-2$. For $X,Y\in\mathcal F_1$, downward closure gives $X\setminus Y,Y\in\mathcal F_1$. Compare $(X\setminus Y)\cup\{n\}$ with $Y\in\mathcal F_0$. Their distance is $|X\cup Y|+1$, so $|X\cup Y|\le d-1<n-1$. Choose $i<n$ outside their union. Shiftedness puts $X\cup\{i\}$ in $\mathcal F_0$, while $Y\cup\{n\}$ is a member of $\mathcal F$. Their distance is $|X\triangle Y|+2$, proving the claim. For $d<2$ this argument forces $\mathcal F_1$ to be empty.
The induction now gives
$$
|\mathcal F|\le D(n-1,d)+D(n-1,d-2)=D(n,d),
$$
where negative-diameter <set families> are empty. The equality of the numerical expressions is Pascal's identity. The base case $d=0$ has at most one member. At the boundary $d=n-1$, at most one member of each complementary pair is allowed, giving $2^{n-1}$, which equals $D(n,n-1)$ by binomial symmetry. For the remaining cases $1\le d\le n-2$, both section bounds belong to smaller-dimensional induction cases. This completes the diametric proof.
For a $t$-intersecting <set family>,
$$
|A\triangle B|=|A\cup B|-|A\cap B|\le n-t=d,
$$
so the theorem applies. For $d=2q$, equality is attained by all <sets> of size at least $n-q$, since two such <sets> intersect in at least $n-2q=t$ elements. For $d=2q+1$, take all <sets> of size at least $n-q$, and also all $(n-q-1)$-sets avoiding one fixed point. Two of these smaller <sets> intersect in at least $2(n-q-1)-(n-1)=t$ points; a smaller and a larger <set> intersect in at least $t$; two larger <sets> intersect in at least $t+1$. The size of this <set family> is
$$
\sum_{i=0}^q\binom ni+\binom{n-1}q
=2\sum_{i=0}^q\binom{n-1}i,
$$
so it attains the odd bound. If $t=0$, the full power <set> is admissible; if $t>n$, the only admissible <set family> is empty.
\b[Every maximal 1-intersecting <set family> attains $2^{n-1}$, for $n\ge1$.] Such a <set family> is an <up-set>, since adding a superset preserves <set intersections>. It contains $[n]$ and omits the empty <set>. It contains at most one member of any complementary pair. If neither of two nonempty complements $A,A^c$ were present, maximality would give a member $B$ disjoint from $A$; upward closure would then include $A^c$, a contradiction. Thus it chooses precisely one member of each complementary pair, as in <maximal intersecting families choose one member of every complementary pair>.
\b[Not every maximal 2-intersecting <set family> attains the bound.] On four points, let $\mathcal A=\{A:\{1,2\}\subseteq A\}$. This <set family> has four members and is maximal: every missing <set> meets its member $\{1,2\}$ in at most one point. The sharp bound is $1+4=5$, attained by all <sets> of size at least three. This proves <maximal intersection does not imply maximum size>.
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