10. Given $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3} \in \mathbf{N}$ and satisfy $a_{1}+a_{2}+a_{3}=b_{1}+b_{2}+b_{3}$
$$
\begin{array}{c}
a_{1} a_{2} a_{3}=b_{1} b_{2} b_{3} \\
a_{1} a_{2}+a_{1} a_{3}+a_{2} a_{3} \neq b_{1} b_{2}+b_{1} b_{3}+b_{2} b_{3}
\end{array}
$$

then the minimum value of the maximum number among $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}$ is $\qquad$
10. 8 .
$a_{1}=2, a_{2}=a_{3}=6, b_{1}=b_{2}=3 \quad b_{3}=8$ satisfies the condition. Let $A=\left\{a_{1}, a_{2}, a_{3}\right\}$, $B=\left\{b_{1}, b_{2}, b_{3}\right\}$. If $\max A \cup B \leqslant 7$, then $A \cap B=\varnothing$. Otherwise, without loss of generality, assume $a_{1}=b_{1}$,
then $\left\{\begin{array}{l}a_{2} a_{3}=b_{2} b_{3} \\ a_{2}+a_{3}=b_{2}+b_{3}\end{array}\right.$ and $a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1} \neq b_{1} b_{2}+b_{2} b_{3}+b_{3} b_{1}$
so $A \cap B=\varnothing$.
Then $5,7 \notin A \cup B$ (because $a_{1} a_{2} a_{3}=b_{1} b_{2} b_{3}$ ),
so $A \cup B \subset\{1,2,4,3,6\}$.
Therefore, $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}$ must have repetitions, so without loss of generality, assume $a_{1}=a_{2}$.
Below, we discuss the cases:
(1) $a_{1}=a_{2}=6$, then without loss of generality, assume $b_{1}=b_{2}=3$,
then $b_{3}=4 a_{3} \Rightarrow b_{3}=4, a_{3}=1$ is not a solution.
(2) $a_{1}=a_{2}=4$, then because $4 \notin B$,
so $2^{4} \times b_{1} b_{2} b_{3}$ contradicts $a_{1} a_{2} a_{3}=b_{1} b_{2} b_{3}$.
(3) $a_{1}=a_{2}=3$. Then without loss of generality, assume $b_{1}=b_{2}=6$, which is the same as $1^{\circ}$.
(4) $a_{1}=a_{2}=2$, then $1 \in B$. Otherwise, because $2 \notin B$,
so $b_{1} b_{2} b_{3} \geqslant 3^{3}>2 \cdot 2 \cdot 6 \geqslant a_{1} a_{2} a_{3}$, which is a contradiction.
So without loss of generality, assume $b_{1}=1$, then $b_{2} b_{3}=4 a_{3}$.
$b_{2}+b_{3}=a_{3}+3$ so $b_{2} b_{3}=4 b_{2}+4 b_{3}-12$
$\Rightarrow\left(b_{2}-4\right)\left(b_{3}-4\right)=4=1 \times 4=2 \times 2=(-4) \times(-1)=(-2) \times(-2)$
get $\left(b_{2}, b_{3}\right)=(5,8),(6,6),(0,3)(2,2)$, none of which fit the condition.
(5) $a_{1}=a_{2}=1$ then $1 \in B, b_{2} \geqslant 2, b_{2} \geqslant 2, b_{3} \geqslant 2$,
and $a_{3}=b_{1} b_{2} b_{3} \geqslant b_{1}+b_{2} b_{3}$ (because $\left.\left(b_{1}-1\right)\left(b_{2} b_{3}-1\right) \geqslant 1\right) \geqslant b_{1}+b_{2}+b_{3}$
because $\left(b_{2}-1\right)\left(b_{3}-1\right) \geqslant 1$, which is a contradiction, so no solution.
In summary, (1), (2), (3), (4), (5) show that $\max A \cup B \leqslant 7$ has no solution, so $(\max A \cup B)_{\text {min }}=8$.