By Euler's totient, since $\varphi(10)\mid 2000$, as long as $a$ is relatively prime to 10, the divisibility is satisfied. Obviously if $a$ isn't relatively prime, the divisibility can't be satisfied.

This is simply a question of "Find all positive integers $a$ such that $a^{2000}$ ends with a 1." In other words, $a^{2000} \equiv 1 \pmod {10}.$ Therefore, $a \equiv x \pmod {10},$ where $0 \ge x \ge 9.$ Next, we can just try all of these. It is quite obvious that $x$ cannot be $0.$ Also, it is clear that $x$ can be 1. It is also clear that no even number can be $x.$ Thus, we need only try 3, 5, and 7 for $x.$ What is $3^{2000} \pmod {10}?$ The pattern goes 3, 9, 7, 1, 3, 9, 7, 1. Thus, as 2000 mod 4 = 0, we have that this is $1.$ What is $5^{2000} \pmod {10}?$ This is obviously 5 What is $7^{2000} \pmod {10}?$ The pattern goes 7, 9, 3, 1, 7, 9, 3, 1. Thus, this is also $1.$ And the value for 9 is also 1, because 2000 is even. thus, $a \equiv 1, 3, 7, 9 \pmod {10}.$