HMMT February

Let $x_{k}$, for $1 \leq k \leq 40$, be the number of integers $i$ with $1 \leq i \leq 20$ such that $a_{i} \geq k$. Let $y_{k}$, for $1 \leq k \leq 40$, be the number of integers $j$ with $1 \leq j \leq 20$ such that $b_{j} \geq k$. It follows from the problem statement that $x_{k}+y_{k}$ is the number of elements of the set $\{1, \ldots, 40\}$ which are greater than or equal to 40, which is just $41-k$. Note that if $1 \leq i, j \leq 20$, and $1 \leq k \leq 40$, then $\min (a_{i}, b_{j}) \geq k$ if and only if $a_{i} \geq k$ and $b_{j} \geq k$. So for a fixed $k$ with $1 \leq k \leq 40$, the number of pairs $(i, j)$ with $1 \leq i, j \leq 20$ such that $\min (a_{i}, b_{j}) \geq k$ is equal to $x_{k} y_{k}$. So we can rewrite $\sum_{i=1}^{20} \sum_{j=1}^{20} \min (a_{i}, b_{j})=\sum_{k=1}^{40} x_{k} y_{k}$. Since $x_{k}+y_{k}=41-k$ for $1 \leq k \leq 40$, we have $x_{k} y_{k} \leq\left\lfloor\frac{41-k}{2}\right\rfloor\left\lceil\frac{41-k}{2}\right\rceil$ by a convexity argument. So $\sum_{i=1}^{20} \sum_{j=1}^{20} \min (a_{i}, b_{j}) \leq \sum_{k=1}^{40}\left\lfloor\frac{41-k}{2}\right\rfloor\left\lceil\frac{41-k}{2}\right\rceil=5530$. Equality holds when $(a_{1}, \ldots, a_{20})=(2,4, \ldots, 38,40)$ and $(b_{1}, \ldots, b_{20})=(1,3, \ldots, 37,39)$.