From the sequence of natural numbers, we remove the perfect squares. In the remaining sequence of numbers, which is the 2001st, and which position does the number 2001 occupy?
Since $44^{2}=19362001$, the last square number we missed was $44^{2}$. This is precisely the 45th omitted number, because 0 is a natural number and a square number as well.

For the second part of the question, we can give the following answer: originally, 2001 was in the 2002nd position in the (starting with 0) sequence. If we omit the 45 square numbers before it in the sequence, it will move to the $2002-45=1957$th position.

Answering the first part of the question: starting from the 1957th position, we need to write down 45 more numbers to reach the 2001st term of the sequence. However, within the $[2001,2045]$ interval, there is a square number, $2025=45^{2}$. We need to omit this, so in the sequence, the 2001st position will be occupied by 2046.

Kovács Orsolya (Szekszárd, I. Béla Gymn. and Inf. Secondary School, 11th grade)