Using duster over and over and over again!

Let's start small . Let the numbers be a , b , c , d . Choosing any two say a and b . Now we erase them and write as a + b + ab = (a+1)(b+1) - 1. Now either you take this number or choose any two number you will observe that what we are basically doing is :

1.Choosing two number .

2.Erasing them from board.

3.Adding 1 to each of them.

4.Miltiplying both the resulting numbers (from operation 3).

5.Subtracting one from the product and then writing it on the board.

On doing it continuously we will observe that :

For $1 , \frac{1}{2} , \frac{1}{3} , \cdots , \frac{1}{n} n \in N .$ We get last number as

$[(1 +1)(1+\frac{1}{2}) \cdots (1+\frac{1}{n} ] - 1$

$= [(2)(\frac{3}{2})(\frac{4}{3})(\frac{5}{4})\cdots(\frac{n+1}{n})] - 1 = (n+1) - 1 = n .$

So in the question asked answer would be $\boxed{16041998}.$