4. Notice that, $\alpha n-1\gamma \geqslant \delta>\beta$.
Thus, there exists a positive integer $N$ satisfying
(1) $(\alpha-\gamma) N>1$;
(2) $(\delta-\beta) N>1$;
(3) $\beta N>1$.

For any $m>N$, the above three conditions hold. At this point,
$[\alpha m]-[\gamma m] \geqslant 1, [\delta m]-[\beta m] \geqslant 1$,
$[\beta m] \geqslant 1$.
First, we prove a lemma.
Lemma When $m>N$, under the above conditions, $\alpha m$,
$\beta m$, $\gamma m$, $\delta m$ are all positive integers.
Proof Fix $m>N$. Let $[\alpha m]=A, \{\alpha m\}=a$, and the binary representation of $a$ is $a=(0.a_1 a_2 \cdots)_2$;
Similarly, we can define $[\beta m]=B, \{\beta m\}=b$,
$b$'s binary representation is $b=(0.b_1 b_2 \cdots)_2$;
$[\gamma m]=C, \{\gamma m\}=c$,
$c$'s binary representation is $c=(0.c_1 c_2 \cdots)_2$;
$[\delta m]=D, \{\delta m\}=d$,
$d$'s binary representation is $d=(0.d_1 d_2 \cdots)_2$.
According to the problem's conditions,
$[\alpha m][\beta m]=[\gamma m][\delta m]$.
Then $A B=C D$, and $A>C \geqslant D>B \geqslant 1$.
Take $n=2 m$, then
$$\begin{array}{l} {[2 \alpha m]=2 A+a_{1}, [2 \beta m]=2 B+b_{1},} \\ {[2 \gamma m]=2 C+c_{1}, [2 \delta m]=2 D+d_{1} \text {. }} \\ \text { By } [2 \alpha m][2 \beta m]=[2 \gamma m][2 \delta m], \text { we have } \\ \left(2 A+a_{1}\right)\left(2 B+b_{1}\right)=\left(2 C+c_{1}\right)\left(2 D+d_{1}\right) \\ \Rightarrow 2 A b_{1}+2 B a_{1}+a_{1} b_{1} \\ \quad=2 C d_{1}+2 D c_{1}+c_{1} d_{1} \text {. } \end{array}$$
(1) When $a_{1}=1, b_{1}=1$, the left side of equation (2) is odd, so the right side is odd. Therefore, $c_{1}=1, d_{1}=1$. From equation (2), we get $A+B=C+D$, and combining $A B=C D$, we have $\{A, B\}=\{C, D\}$, which contradicts $A>C \geqslant D>B \geqslant 1$.
(2) When $a_{1}=1, b_{1}=0$, the left side of equation (2) is even, so the right side is even. Therefore, $c_{1} d_{1}=0$. Equation (2) becomes $B=d_{1} C+c_{1} D \in\{0, C, D\}$. Contradiction.
(3) When $a_{1}=0, b_{1}=1$, the left side of equation (2) is even, so the right side is even. Therefore, $c_{1} d_{1}=0$. Equation (2) becomes $A=d_{1} C+c_{1} D \in\{0, C, D\}$. Contradiction.
(4) When $a_{1}=0, b_{1}=0$, from equation (2) we get
$c_{1}=d_{1}=0$.
Thus, $a_{1}=b_{1}=c_{1}=d_{1}=0$.
Take $n=4 m$, then
$[4 \alpha m]=4 A+a_{2}, [4 \beta m]=4 B+b_{2}$,
$[4 \gamma m]=4 C+c_{2}, [4 \delta m]=4 D+d_{2}$.
By $[4 \alpha m][4 \beta m]=[4 \gamma m][4 \delta m]$, we have
$\left(4 A+a_{2}\right)\left(4 B+b_{2}\right)=\left(4 C+c_{2}\right)\left(4 D+d_{2}\right)$
$\Rightarrow 4 A b_{2}+4 B a_{2}+a_{2} b_{2}=4 C d_{2}+4 D c_{2}+c_{2} d_{2}$.
Similar to the previous discussion, we know
$a_{2}=b_{2}=c_{2}=d_{2}=0$.
Using mathematical induction, we can prove that for any
$k \in \mathbf{N}_{+}, a_{k}=b_{k}=c_{k}=d_{k}=0$.
Therefore, $a=b=c=d=0$, and $\alpha m, \beta m, \gamma m, \delta m$ are all positive integers.
The lemma is proved.
According to the lemma, $(m+1) \alpha, (m+1) \beta, (m+1) \gamma$, $(m+1) \delta$ are all positive integers.
Therefore, $\alpha, \beta, \gamma, \delta$ are all positive integers.