4. (3 points) Warehouses A and B originally each stored whole bags of grain. If 90 bags are transferred from Warehouse A to Warehouse B, then the number of bags in Warehouse B will be twice that in Warehouse A. If a certain number of bags are transferred from Warehouse B to Warehouse A, then the number of bags in Warehouse A will be 6 times that in Warehouse B. How many bags of grain did Warehouse A originally have at least? 

The original number of bags in Warehouse A is $\qquad$.
【Analysis】Two equations are: $(a - 90) \times 2 = b + 90$; $a + c = (b - c) \times 6$, and we need to find the smallest integer solution accordingly.

【Solution】Let the original number of bags of grain in Warehouse A be $a$, and in Warehouse B be $b$. According to the problem, we have $2(a-90)=b+90$ (1);

If Warehouse B transfers $c$ bags to Warehouse A, then the number of bags in Warehouse A is 6 times that in Warehouse B, i.e., $a+c=6(b-c)$ (2);
From (1), we get $b=2a-270$ (3),

Substituting (3) into (2) and simplifying, we get $11a-7c=1620$,
Since $c=\frac{11a-1620}{7}=a-232+\frac{4(a+1)}{7}$
And $a, c$ are positive integers, thus $\frac{11a-1620}{7} \geqslant 1$, which means $a \geqslant 148$;
And 7 divides $4(a+1)$,
Since 4 and 7 are coprime, 7 must divide $a+1$, the smallest value of $a+1$ is 154, so the smallest value of $a$ is 153.
Answer: The original minimum number of bags of grain in Warehouse A is 153.
Therefore, the answer is: 153.