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Prove that there are 2020$2020$ consecutive numbers such that the number of divisors of each number is a multiple of 1399$1399$?

Q. Prove that there are $2020$ consecutive numbers such that the number of divisors of each number is a multiple of $1399$?

Prove that there are 2020 consecutive numbers such that the number of divisors of each number is a multiple of 1399? II thought maybe choosing P1, P2,...P2020$P_1, P_2,\ldots P_{2020}$ as 2020$2020$ distinct primes, we can say there exists n$n$, such that { n=0 (mod P1^1398), n=-1 (mod P2^1398) ... n=-2019 (mod P2020^1398) }${ n=0 (\mod P_1^{1398}), n=-1 (\mod P_2^{1398}),\ldots, n=-2019 (\mod P_{2020}^{1398}) }$ with the Chinese remainder theorem, but I don't think this is right and I don't know the reason. If this statement is true, does it answer the question, and if not, how should I proceed? (Also, 1399$1399$ and 2020$2020$ aren't special numbers, so I suppose 1399$1399$ being a prime is an accident, and I shouldn't focus on it. But I don't know how I would think about the problem if the numbers differed and the number chosen instead of 1399$1399$ was composite.)

Prove that there are 2020 consecutive numbers such that the number of divisors of each number is a multiple of 1399?

Prove that there are 2020 consecutive numbers such that the number of divisors of each number is a multiple of 1399? I thought maybe choosing P1, P2,...P2020 as 2020 distinct primes, we can say there exists n, such that { n=0 (mod P1^1398), n=-1 (mod P2^1398) ... n=-2019 (mod P2020^1398) } with the Chinese remainder theorem, but I don't think this is right and I don't know the reason. If this statement is true, does it answer the question, and if not, how should I proceed? (Also, 1399 and 2020 aren't special numbers, so I suppose 1399 being a prime is an accident, and I shouldn't focus on it. But I don't know how I would think about the problem if the numbers differed and the number chosen instead of 1399 was composite.)

Prove that there are $2020$ consecutive numbers such that the number of divisors of each number is a multiple of $1399$?

Q. Prove that there are $2020$ consecutive numbers such that the number of divisors of each number is a multiple of $1399$?

I thought maybe choosing $P_1, P_2,\ldots P_{2020}$ as $2020$ distinct primes, we can say there exists $n$, such that ${ n=0 (\mod P_1^{1398}), n=-1 (\mod P_2^{1398}),\ldots, n=-2019 (\mod P_{2020}^{1398}) }$ with the Chinese remainder theorem, but I don't think this is right and I don't know the reason. If this statement is true, does it answer the question, and if not, how should I proceed? (Also, $1399$ and $2020$ aren't special numbers, so I suppose $1399$ being a prime is an accident, and I shouldn't focus on it. But I don't know how I would think about the problem if the numbers differed and the number chosen instead of $1399$ was composite.)

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Prove that there are 2020 consecutive numbers such that the number of divisors of each number is a multiple of 1399?

Prove that there are 2020 consecutive numbers such that the number of divisors of each number is a multiple of 1399? I thought maybe choosing P1, P2,...P2020 as 2020 distinct primes, we can say there exists n, such that { n=0 (mod P1^1398), n=-1 (mod P2^1398) ... n=-2019 (mod P2020^1398) } with the Chinese remainder theorem, but I don't think this is right and I don't know the reason. If this statement is true, does it answer the question, and if not, how should I proceed? (Also, 1399 and 2020 aren't special numbers, so I suppose 1399 being a prime is an accident, and I shouldn't focus on it. But I don't know how I would think about the problem if the numbers differed and the number chosen instead of 1399 was composite.)