1. We choose a number uniformly at random from the set $\{1,2,3,\dots,2000\}.$ So total number of possible choices is $2000.$

2. We need the probability that the chosen number is a multiple of $3$ or a multiple of $7$. This is a standard inclusion-exclusion problem.

Let

• $A$ = set of multiples of $3$ from $1$ to $2000$
• $B$ = set of multiples of $7$ from $1$ to $2000$

Then $P(A\cup B)=\frac{n(A)+n(B)-n(A\cap B)}{2000}.$

3. Count multiples of $3$ up to $2000$: $n(A)=\left\lfloor \frac{2000}{3} \right\rfloor=666.$

4. Count multiples of $7$ up to $2000$: $n(B)=\left\lfloor \frac{2000}{7} \right\rfloor=285.$

5. Numbers which are multiples of both $3$ and $7$ are multiples of $\operatorname{lcm}(3,7)=21.$ So $n(A\cap B)=\left\lfloor \frac{2000}{21} \right\rfloor=95.$

6. By inclusion-exclusion, $n(A\cup B)=666+285-95=856.$ Hence $p=\frac{856}{2000}=\frac{107}{250}.$

7. Now compute $500p$: $500p=500\cdot \frac{107}{250}=2\cdot 107=214.$

Therefore, the required integer is $\boxed{214}.$

8. Comparison with stored correct answer:

• Derived answer = $214$
• Stored correct answer = $214$

They match.