We have collected 1001 pebbles into a pile. We arbitrarily divide the pile into two parts and multiply the number of pebbles in each part. Then, we repeat this process on one of the parts - if it contains at least two pebbles - by dividing it into two new parts and calculating the product of the number of pebbles in the two new parts. We continue this process until only one pebble remains in each part.

Between what limits does the sum of the 1000 products obtained this way vary?
I. solution. The sum of the products does not depend on the sequence of divisions carried out during the process. In our case, the result is always $500500$, and in general, for $n$ pebbles, it is $\binom{n}{2}=\frac{n(n-1)}{2}$.

After formulating the conjecture, a proof by complete induction is natural, but we must be careful to interpret the sum in question for a single pebble as $\binom{1}{2}=0$. This corresponds to the fact that a pile with only one pebble cannot be divided further, so in this case, the product sum is 0.

Now let $n>1$ and suppose that the statement is true for all $m$ such that $1 \leq m < n$. Divide the pile of $n$ pebbles into a group of $k$ and a group of $n-k$ pebbles in the first step. ($0<k<n$, but $k=1$ is possible!) The product sum calculated for the $n$ pebbles is then equal to the product obtained from this first division, $k(n-k)$, plus the product sums associated with the groups of $k$ and $n-k$ pebbles. The latter two, according to the induction hypothesis, are $\binom{k}{2}$ and $\binom{n-k}{2}$, independent of further divisions, and the sum of these three quantities is indeed $\binom{n}{2}$ for all $0<k<n$.

We have thus shown that the statement is also true for $n$, and with this, we have completed the proof.

II. solution. We will show that during the process, we are counting exactly the pairs that can be formed from the $n$ pebbles, the number of which, as is well known, is $\binom{n}{2}$. Indeed, the product obtained after the $k$-th division is the number of pairs whose elements were in the same pile until now and are just being placed in different piles. Since we do not combine groups, we count different pairs at each step, and on the other hand, we will get every pair, because by the end of the process, after $n-1$ steps, the piles are singletons, so by then every pair of elements is in different piles.