Integral with Floor

$\begin{aligned} I & = \int_1^{100} x \lfloor x \rfloor \ dx = \sum_{k=1}^{99} \int_k^{k+1} x \lfloor x \rfloor \ dx = \sum_{k=1}^{99} \int_k^{k+1} k x \ dx \\ & = \sum_{k=1}^{99} \frac {kx^2}2 \ \bigg|_k^{k+1} = \sum_{k=1}^{99} \frac k2 \left(k^2 + 2k + 1 - k^2 \right) \\ & = \sum_{k=1}^{99} \left(k^2 + \frac k2 \right) = \frac {99(100)(199)}6 + \frac {99(100)}4 = 330825 \end{aligned}$

Therefore $N \equiv 330825 \equiv \boxed{25} \text{ (mod 100)}$ .

The value of the integral is $\dfrac 12\left (\displaystyle \sum_{i=1}^{99} i(i+1)^2-\displaystyle \sum_{i=1}^{99} i^3\right )$

$=\dfrac 12\displaystyle \sum_{i=1}^{99} (2i^2+i)$

$=330825$

Hence the required answer is $\boxed {25}$ .