A $3,15,24,48, \ldots$ sequence consists of the multiples of 3 that are 1 less than a square number. What is the remainder when the 2001st term of the sequence is divided by 1000?
Let $a_{n}$ denote the $n$-th term of the sequence. For this, it holds that $a_{n}=x^{2}-1$.

If $x$ is a multiple of 3, i.e., $x=3k$, then $a_{n}=x^{2}-1=9k^{2}-1$, which is not a multiple of 3. Therefore, numbers of this form will not appear in the sequence.

For $x=3k+1$, then

$$
a_{n}=(3k+1)^{2}-1=9k^{2}+6k
$$

this is a multiple of 3. Numbers of this form will appear in the sequence.

Finally, if $x=3k+2$, then

$$
a_{n}=(3k+2)^{2}-1=9k^{2}+12k+3,
$$

these will also be terms of the sequence. Thus, we have examined all integers (since an integer is either a multiple of 3, or leaves a remainder of 1 or 2 when divided by 3).

The terms of the sequence are thus $3, 15, 24, 48, 63, 99, \ldots$ It is easy to see that if $n$ is odd, then $k=\frac{n-1}{2}$ and $a_{n}=9k^{2}+12k+3$; for example, for $n=1$, $k=0$ and $a_{1}=3$.

The number 2001 is odd, so $k=\frac{n-1}{2}=\frac{2001-1}{2}=1000$, and since $a_{n}=9k^{2}+12k+3$, $a_{2001}=9 \cdot 10^{6}+12 \cdot 10^{3}+3$, which gives a remainder of 3 when divided by 1000.

Remark. If we increase the terms of the sequence by 1, we get the sequence of square numbers $4, 16, 25, 49, \ldots$; taking the square roots of these, we get the sequence of numbers not divisible by 3, $2, 4, 5, 7, \ldots$. The $k$-th element of this sequence for every odd $k$ is $(k-1) \cdot \frac{3}{2}+2$, and for every even $k$ it is $k \cdot \frac{3}{2}+1$. The 2001st element of the sequence is thus $2000 \cdot \frac{3}{2}+2$ less than the square of $3002$, which is $3002^{2}-1$, and this clearly gives a remainder of $2^{2}-1=3$ when divided by 1000.

The transformed sequence can be briefly written as $b_{k}=1+\left[\frac{3k}{2}\right]$, where $k \geq 1$. For $k=2001$, we again get $b_{k}=1+3001=3002$.