Let $n=2020$. If we let $a=2^{p_{1}} \cdot 3^{q_{1}}, b=2^{p_{2}} \cdot 3^{q_{2}}, c=2^{p_{3}} \cdot 3^{q_{3}}$, then the number of ordered triples $(a, b, c)$ that satisfy the second and third conditions is the number of nonnegative solutions to $p_{1}+p_{2}+p_{3}=n$ and $q_{1}+q_{2}+q_{3}=n$, where at least one of $p_{1}, p_{2}, p_{3}$ is zero and at least one of $q_{1}, q_{2}, q_{3}$ is zero (otherwise, $\operatorname{gcd}(a, b, c)>1$). By complementary counting, the number is $$\left(\binom{n+2}{2}-\binom{n-1}{2}\right)^{2}=9 n^{2}$$ Let $\ell$ be the number of unordered triples $(a, b, c)$ with $a, b, c$ distinct, and $m$ the number of unordered triples $(a, b, c)$ with two numbers equal. Since it is impossible for $a=b=c$, we have $9 n^{2}=6 \ell+3 m$. We now count $m$. Without loss of generality, assume $a=b$. For the factors of 2, we have two choices: either assign $2^{2020}$ to $c$ or assign $2^{1010}$ to both $a$ and $b$. We have a similar two choices for the factors of 3. Therefore $m=4$. Our final answer is $$N=m+n=\frac{6 \ell+3 m+3 m}{6}=\frac{9 \cdot 2020^{2}+12}{6}=2+6 \cdot 1010^{2} \equiv 602 \quad(\bmod 1000)$$