Let integers $l > m > n$ be the side lengths of a triangle satisfying $\left\{\frac{3^l}{10^4}\right\}=\left\{\frac{3^m}{10^4}\right\}=\left\{\frac{3^n}{10^4}\right\}$ where function $\{x\}$ returns the decimal part of real number $x$. Find the least possible value of this triangle's perimeter.
The given condition implies that the last four digits of $3^l$, $3^m$ and $3^n$ are the same. $$\begin{array}{ll} \therefore &10^4\mid (3^m - 3^n) = 3^n\left(3^{m-n}-1\right)\\ \implies & 3^{m-n}\equiv 1\pmod{10^4}\\ \implies& m-n\equiv 0\pmod{500}\end{array}$$
The last conclusion holds because the order of $3$ modulo $10^4$ is $500$. Similarly, we must have $l-m\equiv 0\pmod{500}$ as well.
Let $m-n=500a$ and $l-m=500b$ where $a$ and $b$ are positive integers. By triangular inequality, we have $$n>l-m = 500b \implies n_{min} = 500\times 1 + 1 = 501$$
Accordingly, $m_{min} = 1001$ and $l_{min}=1501$. Therefore, the minimal perimeter equals $501+1001+1501=\boxed{3003}$.