3. The sum of 10 distinct integers is 2012. Then, the minimum value of the difference between the largest and smallest of these 10 integers is ( ).
(A) 10
(B) 11
(C) 12
(D) 13
3. B.

Let these 10 distinct integers be $x_{1}$, $x_{2}, \cdots, x_{10}$, and $x_{1}<x_{2}<\cdots<x_{10}$. Then
$$
\begin{array}{l}
x_{2} \geqslant x_{1}+1, x_{3} \geqslant x_{2}+1 \geqslant x_{1}+2, \\
x_{4} \geqslant x_{1}+3, \cdots, x_{10} \geqslant x_{1}+9, \\
x_{9} \leqslant x_{10}-1, x_{8} \leqslant x_{9}-1 \leqslant x_{10}-2, \\
x_{7} \leqslant x_{10}-3, \cdots, x_{1} \leqslant x_{10}-9 .
\end{array}
$$
$$
\text { Hence } x_{1}+x_{2}+\cdots+x_{10} \geqslant 10 x_{1}+45 \text {, }
$$

and $x_{1}+x_{2}+\cdots+x_{10} \leqslant 10 x_{10}-45$,
i.e., $\left\{\begin{array}{l}2012 \geqslant 10 x_{1}+45, \\ 2012 \leqslant 10 x_{10}-45 .\end{array}\right.$
Solving, we get $x_{1} \leqslant 195 \frac{7}{10}, x_{10} \geqslant 205 \frac{7}{10}$.
When $x_{1}=195, x_{10}=206$, we can take
$$
\begin{array}{l}
x_{2}=197, x_{3}=199, x_{4}=200, x_{5}=201, \\
x_{6}=202, x_{7}=203, x_{8}=204, x_{9}=205 .
\end{array}
$$

Therefore, the minimum value of the difference between the maximum and minimum numbers is
$$
206-195=11 \text {. }
$$