Cono Sur Mathematical Olympiad

Since $a_1<a_2<a_3<\cdots <a_n$ and they are positive integers, we have that $a_i\ge i$ for all $i=1,2,\dots ,n$. In particular, if $n\ge 5$, then $$2023<5^5\le a_n^6<a_1+a_2^2+a_3^3+\cdots +a_n^5\le 2023,$$ which is impossible. Therefore, there are no good lists with $n\ge 5$. Now we analyze the cases where $n\le 4$.

If $n=1$, the only restriction is $a_1\le 2023$ and thus there are 2023 good lists.

If $n=2$, since $a_2^2\le 2023<45^2$ we must have $a_2\le 44$. On the other hand, since $44+44^2=1980<2023$, any choice of two positive integers $a_1$ and $a_2$ such that $a_1<a_2\le 44$ satisfies the conditions. Hence there are $\left(\genfrac{}{}{0ex}{}{44}{2}\right)=946$ good lists with $n=2$.

If $n=3$, we have $a_3^3\le 2023<13^3$ and so $a_3\le 12$. Since $12+12^2+12^3=1884<2023$, it suffices to choose three integers from 1 to 12 to construct our list. Therefore there are $\left(\genfrac{}{}{0ex}{}{12}{3}\right)=220$ good lists with $n=3$.

Finally, if $n=4$, then $a_4^4\le 2023<7^4$, which implies $a_4\le 6$. Since $6+6^2+6^3+6^4=1554<2023$, it suffices to choose four integers from 1 to 6 to construct our list. Consequently, there are $\left(\genfrac{}{}{0ex}{}{6}{4}\right)=15$ good lists with $n=4$.