ZETA_in_olympiad wrote: A function $g: \mathbb{Z} \to \mathbb{Z}$ is called adjective if $g(m)+g(n)>max(m^2,n^2)$ for any pair of integers $m$ and $n$. Let $f$ be an adjective function such that the value of $f(1)+f(2)+\dots+f(30)$ is minimized. Find the smallest possible value of $f(25)$. The function $h(n)=\begin{cases} 128 & 1\leq n\leq 15 \\ n^2-127 & 16\leq n\leq30 \\ n^2+128 & else \end{cases}$ is adjective and satisfies $h(1)+h(2)+\cdots+h(30)=(16^2+17^2+\cdots+30^2)+15$. For any adjective function $f,\pi\in S_{15},n\in\{1,2,\cdots,15\}$ we have $f(n)+f(\pi(n)+15)\geq(\pi(n)+15)^2+1$. Thus \begin{align*} f(1)+f(2)+\cdots+f(30)=\sum_{n=1}^{15}f(n)+f(\pi(n)+15)\geq\sum_{n=1}^{15}(\pi(n)+15)^2+1=15+\sum_{n=16}^{30}n^2 \end{align*}Therefore $h$ is an adjective function such that $h(1)+h(2)+\cdots+h(30)$ is minimized. For any adjective function $f$ with $f(1)+f(2)+\cdots+f(30)=8230$ we must have equality in $f(n)+f(\pi(n)+15)\geq(\pi(n)+15)^2+1$. Thus $f(n)+f(m)=m^2+1$ for $n\in\{1,2,\cdots,15\},m\in\{16,17,\cdots,30\}$ and $f(1)=f(2)=\cdots=f(15)$. Since $f(16)+f(16)>16^2\Rightarrow f(16)\geq129$ we have $f(1)=16^2+1-f(16)\leq128$. Note that $h(25)=25^2+1-128$. If $f(1)<128$, we have $f(1)+f(25)=25^2+1$ and $f(25)>25^2+1-128$. In this case $f(25)$ is not minimal. Thus the smallest possible value of $f(25)$ is $h(25)=25^2+1-128=498$.

@above your solution is remarkable. Here's a notable fact though, you misspelled "function" twice.