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