Chinese Mathematical Olympiad

Proof. Let $S$ be the number of pairs $(i,j)$ that satisfy the following conditions: $$1\le i<j\le 2023,a_i{a}_j\ge 1.$$ Let $T$ be the number of elements among $a_1,a_2,\dots ,a_{2023}$ that are not less than $1$. Then $$100=a_1+a_2+\cdots +a_{2023}\ge T,(3)$$ $$10000=(a_1+a_2+\cdots +a_{2023}{)}^2=\sum _{i=1}^{2023}{a}_i^2+2\sum _{1\le i<j\le 2023}{a}_i{a}_j\ge T+2S.(4)$$ Adding the two equations and dividing by $2$, we get $N=S+T\le 5050$. If equality holds, then equality in the above equation (2) holds, which means each $a_i^2$ is either $0$ or $1$. That is, among $a_1,a_2,\dots ,a_{2023}$, exactly $100$ are $1$s and $1923$ are $0$s. This is a necessary and sufficient condition.