3. Given the set
$$
A=\left\{x \mid x=a_{0}+a_{1} \times 7+a_{2} \times 7^{2}+a_{3} \times 7^{3}\right\} \text {, }
$$

where, $a_{i} \in\{1,2, \cdots, 6\}(i=0,1,2,3)$. If positive integers $m, n \in A$, and $m+n=2014(m>n)$, then the number of pairs of positive integers $(m, n)$ that satisfy the condition is $\qquad$.

3. 551.

Notice that, $2014=5 \times 7^{3}+6 \times 7^{2}+5$.
Given $m, n \in A$, let
$$
\begin{array}{l}
m=a \times 7^{3}+b \times 7^{2}+c \times 7+d, \\
n=a^{\prime} \times 7^{3}+b^{\prime} \times 7^{2}+d^{\prime},
\end{array}
$$

where $a, b, c, d, a^{\prime}, b^{\prime}, c^{\prime}, d^{\prime} \in\{1,2, \cdots, 6\}$.
(1) When $d=1,2,3,4$, $d^{\prime}=5-d$. At this time,
$$
c=1,2, \cdots, 6, c^{\prime}=7-c.
$$

If $b=1,2,3,4$, then $b^{\prime}=5-b$. At this time,
$$
a=1,2,3,4, a^{\prime}=5-a.
$$

If $b=5$, then $b^{\prime} \equiv 0(\bmod 7)$, which cannot be taken.
If $b=6$, then $b^{\prime}=6$. At this time,
$$
a=1,2,3, a^{\prime}=4-a.
$$

Therefore, in this case, the number of $(m, n)$ that meet the conditions is
$$
4 \times 6(4 \times 4+3)=24 \times 19=456 \text{ (pairs). }
$$
(2) When $d=5$, $d^{\prime} \equiv 0(\bmod 7)$, which cannot be taken.
(3) When $d=6$, $d^{\prime}=12-d=6$. At this time,
$$
c=1,2, \cdots, 5, c^{\prime}=6-c.
$$

If $b=1,2,3,4$, then $b^{\prime}=5-b$. At this time,
$$
a=1,2,3,4, a^{\prime}=5-a.
$$

If $b=5$, then $b^{\prime} \equiv 0(\bmod 7)$, which cannot be taken.
If $b=6$, then $b^{\prime}=6$. At this time,
$$
a=1,2,3, a^{\prime}=4-a.
$$

Therefore, in this case, the number of $(m, n)$ that meet the conditions is
$$
5(4 \times 4+3)=95 \text{ (pairs). }
$$

In summary, there are $456+95=551$ pairs of $(m, n)$ that satisfy the conditions.