3. (50 points) Find the smallest prime $p$ that satisfies $(p, N)=1$, where $N$ is the number of all $\left(a_{0}, a_{1}, \cdots, a_{2012}\right)$ that meet the following conditions:
(1) $\left(a_{0}, a_{1}, \cdots, a_{2012}\right)$ is a permutation of $0,1, \cdots, 2012$;
(2) For any positive divisor $m$ of 2013 and all $n(0 \leqslant n<n+m \leqslant 2012)$, $m \mid\left(a_{n+m}-a_{n}\right)$.
3. From (2), we know that for any $i, j (0 \leqslant i < j \leqslant 2012)$ and $m=3^{\alpha} \times 11^{\beta} \times 61^{\gamma} (\alpha, \beta, \gamma \in \{0,1\})$, we have
$$
a_{i} \equiv a_{j}(\bmod m) \Leftrightarrow i \equiv j(\bmod m).
$$

Let
$$
\begin{array}{l}
A_{i}=\{x \in \mathrm{N} \mid 0 \leqslant x \leqslant 2012, x \equiv i(\bmod 3)\}(i=0,1,2), \\
A_{i j}=\left\{x \in A_{i} \mid x \equiv j(\bmod 11)\right\}(j=0,1, \cdots, 10), \\
A_{i j k}=\left\{x \in A_{i j} \mid x \equiv k(\bmod 61)\right\}(k=0,1, \cdots, 60).
\end{array}
$$

From (1), we know that for $i=0,1,2$, all $a_{n} (n \in A_{i})$ are congruent modulo 3 (the remainder is denoted as $i^{\prime}$), and there is a one-to-one mapping
$$
f_{1}: A_{i} \rightarrow A_{i^{\prime}}.
$$

Similarly, there are one-to-one mappings
$$
f_{2}: A_{i j} \rightarrow A_{i^{\prime} j^{\prime}}, f_{3}: A_{i j k} \rightarrow A_{i^{\prime} j^{\prime} k^{\prime}}.
$$

Therefore, there are $3! \times 11! \times 61!$ permutations that satisfy the conditions.
Below is a method to construct all such permutations.
Let the bijection $f: (i, j, k) \rightarrow (i^{\prime}, j^{\prime}, k^{\prime})$, where
$$
0 \leqslant i, i^{\prime} \leqslant 2, 0 \leqslant j, j^{\prime} \leqslant 10, 0 \leqslant k, k^{\prime} \leqslant 60.
$$

If $n$ satisfies
$$
\left\{\begin{array}{l}
n \equiv i(\bmod 3), \\
n \equiv j(\bmod 11), \\
n \equiv k(\bmod 61),
\end{array}\right.
$$

then take $a_{n}$ to be the number that satisfies
$$
\left\{\begin{array}{l}
a_{n} \equiv i^{\prime}(\bmod 3), \\
a_{n} \equiv j^{\prime}(\bmod 11), \\
a_{n} \equiv k^{\prime}(\bmod 61).
\end{array}\right.
$$

By the Chinese Remainder Theorem, the solution is unique.
Clearly, $\left(a_{0}, a_{1}, \cdots, a_{2012}\right)$ satisfies conditions (1) and (2).
Thus, $N=3! \times 11! \times 61!$.
Therefore, $p=67$.