jmc

Let $N$ denote the large positive integer that everyone is discussing.

The two incorrect numbers are consecutive numbers. To get the smallest possible value of $N$, we must maximize the incorrect numbers. As such, we should start with the highest possible incorrect numbers and work down.

Suppose the two incorrect numbers are 24 and 25. Then $N$ must still be divisible by $1,2,3,\dots ,23.$ This means $N$ is divisible by 3 and 8, so $N$ is divisible by $3\cdot 8=24$, contradiction. So the two incorrect numbers cannot be 24 and 25. We can eliminate the other high cases similarly.

One of the incorrect numbers cannot be 22, because $N$ would still be divisible by 2 and 11.

One of the incorrect numbers cannot be 20, because $N$ would still be divisible by 4 and 5.

One of the incorrect numbers cannot be 18, because $N$ would still be divisible by 2 and 9.

On the other hand, suppose the incorrect numbers were 16 and 17. Then $N$ would still be divisible by $1,2,3,\dots ,15,18,19,\dots ,25$. The lcm of these remaining numbers is $$2^3\cdot 3^2\cdot 5^2\cdot 7\cdot 11\cdot 13\cdot 19\cdot 23=787386600,$$which is not divisible by 16 or 17. Thus, the incorrect numbers can be 16 and 17, and the smallest possible value of $N$ is $\boxed{787386600}$.