= Solution
Take the natural parameter range $1\leq t\leq r\leq n$ and admissible indices $0\leq i\leq r-t$. Larger indices give empty <set families>. Write $q=r-t+1$ and <set> $n_0=\infty$ when discussing the first interval. For the nontrivial interval calculation, $t>1$ and $n>2r-t+1$; all prefixes used below then lie inside the ground <set>.
To compare consecutive <complete-intersection candidate families>, put $a=t+2i$ and $b=t+i$. A <set> in $\mathcal A_{i+1}\setminus\mathcal A_i$ has exactly $b-1$ points in $[a]$ and both of the next two points. A <set> in $\mathcal A_i\setminus\mathcal A_{i+1}$ has exactly $b$ points in $[a]$ and neither next point. Therefore
$$
G_i=|\mathcal A_{i+1}\setminus\mathcal A_i|
=\binom{t+2i}{t+i-1}\binom{n-t-2i-2}{r-t-i-1},
$$
$$
L_i=|\mathcal A_i\setminus\mathcal A_{i+1}|
=\binom{t+2i}{t+i}\binom{n-t-2i-2}{r-t-i}.
$$
For $0\leq i<r-t$ these quantities are positive in the present range, and cancellation of <binomial coefficients> gives
$$
\frac{G_i}{L_i}
=\frac{(t+i)(r-t-i)}{(i+1)(n-r-i-1)}.
$$
The numerator minus the denominator is
$$
(i+1)(n_{i+1}-n),\qquad
n_{i+1}=q\left(2+\frac{t-1}{i+1}\right).
$$
Thus $|\mathcal A_{i+1}|$ is larger, equal or smaller than $|\mathcal A_i|$ according as $n$ is smaller, equal or larger than $n_{i+1}$. Since these thresholds decrease with the index, $n_{k+1}<n<n_k$ implies strict increase up to $\mathcal A_k$ and strict decrease thereafter:
$$
\boxed{\max_i|\mathcal A_i|=|\mathcal A_k|.}
$$
At an interior boundary $n=n_{k+1}$, the two consecutive maxima tie. The interval assertion concerns admissible $k$; taking $k>r-t$ would instead refer to an empty <set family> and is not a valid interpretation.
The <Ahlswede-Khachatrian theorem>, in its <cardinality> form, states
$$
\boxed{M(n,r,t)=\max_{0\leq i\leq r-t}|\mathcal A_i|.}
$$
Here $\mathcal A_i$ is interpreted exactly by its intersection with $[t+2i]$, even if that prefix extends beyond $[n]$. Its size can always be written with $m_i=\min\{n,t+2i\}$ as
$$
|\mathcal A_i|=\sum_{j=t+i}^{r}\binom{m_i}{j}\binom{n-m_i}{r-j},
$$
where impossible <binomial coefficients> are zero. When $n\leq2r-t$, the whole level is $t$-intersecting and the formula gives $M=\binom nr$. The nontrivial case has $n>2r-t$.
First, each candidate is a <t-intersecting family>: two members contain at least $2(t+i)-(t+2i)=t$ common points in the prefix. This proves the lower bound for $M$. We now prove the upper bound, using only the compression and generating-family facts expressly allowed in the question.
The first auxiliary fact is that the usual left <coordinate shifts of a set family> preserve <cardinality> and $t$-intersection. Iterating them produces a <left-compressed set family> of the same size. The shift $S_{ab}$ with $a<b$ replaces $b$ by $a$ only when $b$ is present, $a$ is absent, and the replacement is not already in the <set family>. A <set family> is left-compressed when it is fixed by every such shift. Right compression is the reflected notion.
For the second auxiliary fact, a <generating family for a uniform set family> $\mathcal G$ means a collection of <sets> of size at most $r$ with
$$
\mathcal F=\{A\in[n]^{(r)}:G\subseteq A\text{ for some }G\in\mathcal G\}.
$$
The permitted <small-support generating lemma for extremal intersecting families> is this precise statement: if $\mathcal F$ is a maximum-cardinality left-compressed $t$-intersecting <set family>, $n>2r-t$, and
$$
n>(r-t+1)\left(2+\frac{t-1}{j+1}\right)
\quad(j\geq0),
$$
then it has a generating <set family> supported on $[t+2j]$. Reflection gives the same assertion for a right-compressed extremal <set family>, with the supporting interval at the right end. These two auxiliary results are being used as permitted lemmas; the extremal conclusion of the theorem is established in the remaining argument.
Choose a maximum left-compressed <set family> $\mathcal F$. Its complement <set family>
$$
\mathcal F^*=\{[n]\setminus A:A\in\mathcal F\}
$$
is right-compressed, has uniformity $r'=n-r$ and intersection parameter $t'=n-2r+t$, and is extremal for those parameters. Indeed
$$
|([n]\setminus A)\cap([n]\setminus B)|=n-2r+|A\cap B|,
$$
so complementation gives a cardinality-preserving bijection between the two classes of intersecting <set families>. Notice that $r'-t'+1=q$ too.
We need a compatibility observation, which we prove. If $G$ generates $\mathcal F$ and $H$ generates $\mathcal F^*$, then
$$
\boxed{|G\cup H|\geq n-r+t.}
$$
Otherwise extend $G\cup H$ to a <set> $T$ of size $n-r+t-1$. This size is at least both $r$ and $n-r$ because $n\geq2r-t+1$ and $t\geq1$. Choose an $r$-set $A\subseteq T$ containing $G$, and an $(n-r)$-set $B\subseteq T$ containing $H$. Their generating properties imply $A,[n]\setminus B\in\mathcal F$, but
$$
|A\cap([n]\setminus B)|=|A\cup B|-|B|
\leq(n-r+t-1)-(n-r)=t-1,
$$
a contradiction. This <compatibility bound for complementary generating families> is the link between the two compression directions.
Suppose first that $t>1$ and $n>2r-t+1$ lies strictly between $n_{k+1}$ and $n_k$. Put $m=t+2k$ and $h=r-t-k$. The small-support lemma gives generators $\mathcal G$ for $\mathcal F$ supported on $[m]$. It also gives generators $\mathcal H$ for $\mathcal F^*$ supported on $[m+1,n]$. For completeness, the dual inequality required here is
$$
n>q\left(2+\frac{t'-1}{h+1}\right).
$$
Writing $D=n-2q$, it is equivalent to $kD<q(t-1)$, exactly the upper interval inequality $n<n_k$; for $k=0$ it follows from $t>1$. The dual support length is $t'+2h=n-m$.
Either every $G\in\mathcal G$ has $|G|\geq t+k$, or every $H\in\mathcal H$ has $|H|\geq n-r-k$. If neither assertion held, a pair of smaller generators would have union size at most
$$
(t+k-1)+(n-r-k-1)=n-r+t-2,
$$
contradicting the compatibility bound. In the first alternative, every member of $\mathcal F$ has at least $t+k$ points in $[m]$, so $\mathcal F\subseteq\mathcal A_k$. In the second alternative, each complement has at least $n-r-k$ points in $[m+1,n]$, leaving its original <set> with at most $r-t-k$ points there; again $\mathcal F\subseteq\mathcal A_k$. Therefore $M\leq|\mathcal A_k|$, proving the theorem in every strict interval.
At an interior boundary $n=n_{k+1}$, with $0\leq k\leq r-t-1$, apply the small-support lemma at index $k+1$ to the original <set family>, since $n>n_{k+2}$. Its generators are supported on $[t+2k+2]$. The dual generators are still supported on $[t+2k+1,n]$, by the same dual calculation with $h=r-t-k$. Either all original generators have size at least $t+k+1$, or all dual generators have size at least $n-r-k$: failure of both would give union size at most $n-r+t-1$, again impossible. The first alternative places $\mathcal F$ inside $\mathcal A_{k+1}$, and the second places it inside $\mathcal A_k$. The comparison already proved gives $|\mathcal A_k|=|\mathcal A_{k+1}|$, so the common size is $M$.
It remains to cover the endpoint cases, not to assume them implicitly. For $t=1$ and $n>2r$, the small-support lemma with $j=0$ puts an extremal left-compressed <set family>'s generators inside $\{1\}$. Its members must all contain one, giving $M\leq\binom{n-1}{r-1}$, attained by $\mathcal A_0$. For $t=1$, $n=2r$, at most one member of each complementary pair can occur, so $M\leq\tfrac12\binom{2r}{r}=\binom{2r-1}{r-1}$, again attained by $\mathcal A_0$. If $n<2r$, the whole level is intersecting.
For $t>1$ and $n=2r-t+1$, complementation changes the problem into a one-intersection problem with uniformity $q=r-t+1$. Since $n=2q+t-1>2q$, the preceding case gives
$$
M(n,r,t)=\binom{n-1}{q-1}=\binom{n-1}{r},
$$
attained by $\mathcal A_{r-t}$, the <set family> of $r$-sets avoiding the final point. For $n\leq2r-t$, every two $r$-sets meet in at least $2r-n\geq t$, so $M=\binom nr$ and $\mathcal A_{r-t}$ is the whole level. Finally, $t=r$ gives $M=1$ directly. These cases, the strict intervals and their boundaries cover all $1\leq t\leq r\leq n$, completing the proof of the stated maximum formula.
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