Sum-of-floor equation.

We form the following intervals: $I_1=[0,\frac{1}{4}),I_2=[\frac{1}{4},\frac{1}{3}),I_3=[\frac{1}{3},\frac{1}{2}),I_4=[\frac{1}{2},\frac{2}{3}), I_5=[\frac{2}{3},\frac{3}{4}), I_6=[\frac{3}{4},1)$ . The function $f(x)=\lfloor x\rfloor +\lfloor 2x\rfloor +\lfloor 3x\rfloor +\lfloor 4x\rfloor$ is constant on any of these intervals, and its corresponding values would be 0, 1, 2, 4, 5, and 6, respectively. Besides that, it easy to see that $f(x+1)=f(x)+10.$ Therefore, the only natural values of $n$ less than or equal to 99 for which the equation has not solution are: 3, 7, 8, 9, 13,17, 18, 19, 23,27, 28, 29, 33, 37, 38,39,..., 93, 97, 98, 99. Since the sum of $3+7+8+9=27,$ then the sum of the previous numbers is $27+(27+40)+(27+2*40)+(27+3*40)+ ...+ (27+9*(40))= 270 +40(1+2+3+...+9)=270+ 40*\frac{10*9}{2}=\boxed{2070}$