1. Given a finite set of cards. On each of them, either the number 1 or the number -1 is written (exactly one number on each card), and there are 100 more cards with -1 than cards with 1. If for each pair of different cards, the product of the numbers on them is found, and all these products are summed, the result is 1000. How many cards have the number 1?
Answer: 3950.

Solution: Let the number of cards with 1 be $m$, and the number of cards with -1 be $k$. Then, among all pairs, there are $\frac{m(m-1)}{2}$ pairs of two 1s, $\frac{k(k-1)}{2}$ pairs of two -1s, and $m k$ pairs of 1 and -1. Therefore, the sum in the condition is $\frac{m(m-1)}{2}+\frac{k(k-1)}{2}-m k$, from which we get $m^{2}-2 m k+k^{2}=2000+m+k, (m-k)^{2}=2000+m+k$. According to the condition $k-m=100$, so $m+k=8000$ and $2 m=7900, m=3950$.

Comment: The correct relationship between $k$ and $m$ (in addition to $k-m=100$) has been obtained, but without further progress - 3 points.