75th Romanian Mathematical Olympiad

Let $0<a_1<a_2<a_3<\cdots <a_{40}$ so that $a_1+a_2+a_3+\cdots +a_{40}=2025$. Then $a_2\ge a_1+1$, $a_3\ge a_2+1$, $a_4\ge a_3+1$, $\dots$, $a_{40}\ge a_{39}+1$. Consequently, $a_{40}\ge a_1+39\ge a_2+38\ge \cdots \ge a_{38}+2$. Then $40\cdot a_{40}\ge (a_1+39)+(a_2+38)+\cdots +(a_{37}+3)+(a_{38}+2)+(a_{39}+1)+a_{40}$, hence $40\cdot a_{40}\ge 2025+(1+2+\cdots +39)=2805$. Since $a_{40}$ is a positive integer, it follows that $a_{40}\ge 71$. The value $a_{40}=71$ can be achieved: $1+29+34+35+36+\cdots +70+71=2025$, therefore the required minimum is $71$.