8. It is known that Team A and Team B each have several people. If 90 people are transferred from Team A to Team B, then the total number of people in Team B will be twice that of Team A; if some people are transferred from Team B to Team A, then the total number of people in Team A will be 6 times that of Team B. Then, the original minimum number of people in Team A is.
8. 153.

Let the original number of people in team A and team B be $a$ and $b$ respectively.
Then $2(a-90)=b+90$.
Suppose $c$ people are transferred from team B to team A. Then $a+c=6(b-c)$.
From equations (1) and (2), eliminating $b$ and simplifying, we get
$$
\begin{array}{l}
11 a-7 c=1620 \\
\Rightarrow c=\frac{11 a-1620}{7}=a-232+\frac{4(a+1)}{7} \text {. }
\end{array}
$$

Since $a$ and $c$ are positive integers, thus,
$$
c=\frac{11 a-1620}{7} \geqslant 1 \Rightarrow a \geqslant 148 \text {. }
$$

Also, $7 \mid 4(a+1),(4,7)=1$, hence, $7 \mid (a+1)$.
Therefore, the minimum value of $a$ is 153.
Thus, the original number of people in team A is at least 153.