14. 9000 students participate in a math competition. Their exam numbers are $1000,1001,1002, \cdots 9999$. Xiao Ming finds that his exam number is 8210, and his friend Xiao Qiang's exam number is 2180. Their exam numbers are composed of the same digits (in a different order) and the difference is a multiple of 2010. Therefore, the number of such exam numbers (composed of the same digits and the difference is a multiple of 2010) is $\qquad$ pairs.
【Solution】Solution: Let $\overline{a b c d}$ and $\overline{e f g h}$ be composed of the same digits, and $2010 \mid(\overline{a b c d}-\overline{\text { efgh }})$. Since the sum of the digits of $\overline{a b c d}$ and $\overline{e f g h}$ is the same, they have the same remainder when divided by 9, i.e., $9 \mid(\overline{a b c d}-\overline{e f g h})$, thus $6030 \mid(\overline{a b c d}-\overline{\text { efgh }})$.

Considering $0<\overline{a b c d}-\overline{\text { efgh }}<9000$, we have $\overline{a b c d}-\overline{\text { efgh }}=6030, \overline{a b c d}=\overline{\text { efgh }}+6030$.
From the last digit, we know $d=h, \overline{\mathrm{abc}}-603=\overline{\mathrm{efg}}$.
If $c \geqslant 3, \overline{\mathrm{abc}}-603=\overline{(\mathrm{a}-6) \mathrm{b}(\mathrm{c}-3)}$, but $(a-6)+b+(c-3)=a+b+c-9 \neq a+b+c$, $\overline{(a-6) b(c-3)} \neq \overline{e f g}, \overline{a b c}-603=\overline{(a-6) b(c-3)}$, which is not valid.
If $c \leqslant 2, b=0, \overline{\mathrm{abc}}-603=\overline{\mathrm{a} 0 \mathrm{c}}-603=\overline{(\mathrm{a}-7) 9(\mathrm{c}+7)}$, similarly, this situation is also not valid.
Therefore, $b \geqslant 1$.
$\overline{\mathrm{abc}}-603=\overline{(\mathrm{a}-6)(\mathrm{b}-1)(\mathrm{c}+7)} . c+7$ here can be equal to $a$ or $b$. If $a=c+7$, then $b=c+1$, at this time $(a, b, c)$ can be $(7,1,0), (8,2,1)$, and $(9,3,2)$; If $b=c+7$, then $a=c+6$, at this time $(a, b, c)$ can be $(7,8,1)$ and $(8,9,2) .(a, b, c)$ determined, then consider $d, d$ can be any digit from $0,1,2, \cdots 9$. This way, we can get 50 different $a b c d$, and thus the corresponding efgh. Therefore, there are 50 pairs of such numbers, composed of the same digits, and with a difference that is a multiple of 2010.

The answer is 50.