2. Let $c(X)$ denote the number of subsets of set $X$. If $n$ sets $A_{1}, A_{2}, \cdots, A_{n}$ with different numbers of elements satisfy:
$$
\left(\left|A_{1}\right|+1\right)\left(\left|A_{2}\right|+1\right) \cdots\left(\left|A_{n}\right|+1\right)=2006
$$

and
$$
\begin{array}{l}
c\left(A_{1}\right)+c\left(A_{2}\right)+\cdots+c\left(A_{n}\right) \\
=c\left(A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right)+ \\
\quad c\left(A_{1} \cap A_{2} \cap \cdots \cap A_{n}\right)(n \geqslant 2),
\end{array}
$$

then the minimum value of $\max \left\{\left|A_{1}\right|,\left|A_{2}\right|, \cdots,\left|A_{n}\right|\right\}$ is
$\qquad$
2.58.

Assume $\left|A_{1}\right|>\left|A_{2}\right|>\cdots>\left|A_{n}\right| \geqslant 0$.
When $n \geqslant 3$, by $c(X)=2^{|X|}$ and the condition, we know that $2^{\left|A_{1}\right|}+2^{\left|A_{2}\right|}+\cdots+2^{\left|A_{n}\right|}=2^{\left|A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right|}+2^{\left|A_{1} \cap A_{2} \cap \cdots \cap A_{n}\right|}$.
The left side of the above equation is greater than $2^{\left|A_{1}\right|}+2^{\left|A_{n}\right|}$; the right side is less than or equal to $2^{\left|A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right|}+2^{\left|A_{n}\right|}$, hence $\left|A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right| \geqslant\left|A_{1}\right|+1$.
Also, the left side of the above equation is less than $2^{\left|A_{1}\right|+1}$, so $\left|A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right|b)$, then $(a+1)(b+1)=2006=2 \times 17 \times 59=34 \times 59$. Therefore, $a \geqslant 59-1=58$.
When $a=58, b=33$, the minimum value is achieved.