II. Let the three sides of a triangle be integers $l$, $m$, and $n$, and $l > m > n$. It is known that $\left\{\frac{3^{l}}{10^{4}}\right\}=\left\{\frac{3^{m}}{10^{4}}\right\}=\left\{\frac{3^{n}}{10^{4}}\right\}$, where $\{x\}=x-[x]$, and $[x]$ represents the greatest integer not exceeding $x$. Find the minimum perimeter of such a triangle.
3003
II. 【Analysis and Solution】From the given condition, we know that $\frac{3^{l}}{10^{4}}-\left[\frac{3^{l}}{10^{4}}\right]=\frac{3^{m}}{10^{4}}-\left[\frac{3^{m}}{10^{4}}\right]$
$$
=\frac{3^{n}}{10^{4}}-\left[\frac{3^{n}}{10^{4}}\right]
$$

Thus, $3^{l} \equiv 3^{m} \equiv 3^{n}\left(\bmod 10^{4}\right)$
$$
\left\{\begin{array}{l}
3^{l} \equiv 3^{m} \equiv 3^{n}\left(\bmod 2^{4}\right) \\
3^{l} \equiv 3^{m} \equiv 3^{n}\left(\bmod 5^{4}\right)
\end{array}\right.
$$

Since $(3,2)=(3,5)=1$,
$\therefore$ From the above, we know $3^{l-n} \equiv 3^{m-n} \equiv 1\left(\bmod 2^{4}\right)$.
Now let $u$ be the smallest positive integer satisfying $3^{u} \equiv 1\left(\bmod 2^{4}\right)$, then for any positive integer $v$ satisfying $3^{v} \equiv 1\left(\bmod 2^{4}\right)$, we have $u \mid v$, i.e., $u$ divides $v$. In fact, if $u \mid v$, then by the division algorithm, there exist non-negative integers $a$ and $b$, such that $v=a u$ $+b$, where $0<m>n$, so we have $r>s$.
In this way, the three sides of the triangle are $500 r+n$, $500 s+n$, and $n$. Since the difference between two sides is less than the third side, we have $n>$ $500(r-s)$, therefore, when $s=1, r=2, n=501$, the perimeter of the triangle is minimized, with a value of $(1000+501)+(500+$ $501)+501=3003$.