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$P(x)=A\displaystyle\prod_{i=1}^{99}(x-i)+x\cdot 99!$

Using $P(100)=0$ , we get $A=-100$ .

Hence, $\large P(x)=-100\displaystyle\prod_{i=1}^{99}(x-i)+x\cdot 99!$

Coef of $x^{98}$ in P(x):

$-100\underbrace{(-1+(-2)+(-3)+\cdots+(-99))}_{-\frac{(99)(100)}2}$

$\large =100\times 99\times 50=\boxed{495000}$

Define a polynomial $f(x)$ of degree $99$ such that $f(x)=P(x)-x \cdot 99!$ . We have $f(x)=0$ for $x=1,2,3,...,99$ . By Remainder-Factor Theorem, we have $f(x)=A(x-1)(x-2)(x-3)...(x-99)$ for some constant $A$ . For $x=100$ , we must have $f(x)=-100!$ , therefore $A=-100$ . Note that $f(x)$ and $P(x)$ have the same coefficient for $x^{98}$ which is equal to $-100$ times the negative value of the sum of the roots of $f(x)$ .

$(-100)(-1)(1+2+3+...+99)= \boxed {495000}$