We have that $2^8\cdot3\cdot5>2020$, so $N$ is divisible only by $2$ and $3$ and it should be easy to look at all the cases from here. @below Yeah, this was just off the top of my head.

Math00954 wrote: We have that $2^8\cdot3\cdot5>2020$, so $N$ is divisible only by $2$ and $3$ and it should be easy to look at all the cases from here. The prime factorization does not necessarily have to contain all of the first $k$ primes. $2^{7-n}5^n$ works for $1\le n\le 3$ and $2^8\cdot 7$ does as well.

I proposed this problem. This is more or less the solution I sent First, let's prove that $k\leq 3$. If there are at least $4$ prime numbers, we have that $p_1 + p_2 + p_3 + p_4 \geq 2 + 3 +5 +7 = 17$. $x_1 + x_2 + x_3 + x_4 = p_1 + p_2 + p_3 + p_4 \geq 17 \rightarrow N > 2^{x_1}\cdot 2^{x_2} \cdot 2^{x_3} \cdot 2^{x_4} = 2^{x_1+x_2+x_3+x_4} = 2^{17} > 2020$, which is a contradiction. Now we have three cases: $k = 3$ Assume WLOG that $p_1 < p_2 < p_3$. $N = p_{1}^{x_1} \cdot p_{2}^{x_2} \cdot p_{3}^{x_3} > 2^{x_1}\cdot 2^{x_2} \cdot 4^{x_3} = 2^{x_1+x_2+2x_3} = 2^{p_1 + p_2 + p_3 + x_3} \geq 2^{2+3+5+x_3} = 2^{10 + x_3} \geq 2^{11} > 2020$. We have a contradiction, so there are not solutions for $k = 3$. $k = 2$ We have that $p_1 + p_2 = x_1 + x_2 \leq 10$. We are limited to use the prime numbers $2, 3, 5, 7$. We have to do some casework with the pairs $(2,3), (2,5), (2,7), (3,5), (3, 7)$ Case 1: $x_1+x_2 = 2+3=5$ We have the solutions $N = 3^4\cdot 2, 3^3\cdot 2^2, 3^2\cdot 2^3, 3\cdot 2^4$ Case 2: $x_1+x_2 = 2+5=7$ $5^4\cdot 2^3 > 2020$, the solutions for this case are $N = 5^3 \cdot 2^4, 5^2 \cdot 2^5, 5\cdot 2^6$ Case 3: $x_1+x_2 = 2 + 7$ The only solution is $N = 2^8\cdot 7$ Case 4:$x_1+x_2 = 3+5 = 8$ Since $3^7 \cdot 5 > 2020$, there are not solutions in this case. Case 5:$x_1+x_2 =3+7=10$ Since $3^9\cdot 7 > 2020$, there are not solutions in this case. $k = 1$ We take $N = 2^2, 3^3$. The interoceanic numbers are: $4, 27, 48, 72, 108, 162, 320, 800, 1792, 2000$