20. Express the integer 2013 as
$$
2013=\frac{a_{1}!\cdot a_{2}!\cdots \cdots a_{m}!}{b_{1}!\cdot b_{2}!\cdots \cdots \cdot b_{n}!},
$$

where, $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{m}, b_{1} \geqslant b_{2} \geqslant \cdots \geqslant b_{n}, a_{i} 、 b_{j}$ $\in \mathbf{Z}_{+}(i=1,2, \cdots, m, j=1,2, \cdots, n)$.

When $a_{1}+b_{1}$ takes the minimum value, $\left|a_{1}-b_{1}\right|=$ ( ).
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
20. B.

Given $2013=61 \times 11 \times 3$, and 61 is in the numerator, hence $a_{1}=61$.

Notice that, there does not exist a prime $p$ such that: $p$ is not a prime factor of 2013, $b_{1}<p<61$.

Since primes represent the numerator, not the denominator, the largest $p<61$ is 59, so 59 must be in the numerator, $b_{1}=59$. Therefore,
$$
\begin{array}{l}
2013=\frac{61!\times 19!\times 11!}{59!\times 20!\times 10!} . \\
\text { Thus, }\left|a_{1}-b_{1}\right|=|61-59|=2 .
\end{array}
$$