In the integer sequence $\left\{a_{n}\right\}$, each term takes an integer between $0 \sim 2013$, and from the fourth term onwards, each term is equal to the remainder of the sum of its previous three terms divided by 2014. Find all possible values of the smallest positive period of the sequence $\left\{a_{n}\right\}$.
It is known that the characteristic equation of the sequence $\{a_n\}$ is
$$
\lambda^3 - \lambda^2 - \lambda - 1 = 0 \quad (\text{mod } 2014),
$$
with 13 being one of its roots.
Factoring, we get
$$
(\lambda - 13)(\lambda^2 + 12\lambda + 155) = 0.
$$

Consider the cases modulo 2, 19, and 53 separately.
(1) Clearly, the smallest positive period modulo 2 can be 1, 2, or 4.
(2) Under modulo 19, the three characteristic roots are 13 (i.e., -6), -6 + 9i, and -6 - 9i. Thus,
$$
a_n = p(-6)^n + q(-6 + 9i)^n + r(-6 - 9i)^n,
$$
where all operations are modulo 19. -6 is a primitive root modulo 19, and -6 + 9i is a "super primitive root" (order $19^2 - 1 = 360$).
[Note] In fact,
$$
\begin{array}{l}
(-6 + 9i)^5 = -6 - 6i, \\
(-6 + 9i)^{10} = -4i, \\
(-6 + 9i)^{20} = 3,
\end{array}
$$
where 3 is a primitive root modulo 19.
Therefore, the smallest positive period of the sequence $\{a_n\}$ modulo 19 can be 1, 18, or 360 (1 when $p = q = r = 0$, 18 when $q = r = 0$ but $p \neq 0$, and 360 when $q \neq 0$ or $r \neq 0$).
(3) Under modulo 53, the three characteristic roots are 13, 20, and 21. Thus,
$$
a_n = p \cdot 13^n + q \cdot 20^n + r \cdot 21^n,
$$
where all operations are modulo 53. The order of 13 modulo 53 is 13, and both 20 and 21 are primitive roots modulo 53. Therefore, the smallest positive period of the sequence $\{a_n\}$ modulo 53 can be 1, 13, or 52.

Taking one number from each of the sets $\{1, 2, 4\}$, $\{1, 18, 360\}$, and $\{1, 13, 52\}$ and calculating their least common multiple, the set of answers is
$$
\{1, 2, 4, 13, 18, 26, 36, 52, 234, 360, 468, 4680\}.
$$

Since 2, 19, and 53 are all prime numbers, any combination is possible.