jmc

Let $n=\lfloor x\rfloor ,$ and let $\{x\}=(0.x_1{x}_2{x}_3{x}_4\dots {)}_{10},$ so the $x_i$ are the decimal digits. Then the given condition becomes $$\lfloor y\rfloor \le 100\{x\}-\lfloor x\rfloor =(x_1{x}_2.x_3{x}_4\dots {)}_{10}-n.$$Since $\lfloor y\rfloor$ is an integer, this is equivalent to $$\lfloor y\rfloor \le (x_1{x}_2{)}_{10}-n.$$First, let's look at the interval where $0\le x<1,$ so $n=0.$ For $0\le x<0.01,$ we want $$\lfloor y\rfloor \le 0,$$so $0\le y<1.$

For $0.01\le x<0.02,$ we want $$\lfloor y\rfloor \le 1,$$so $0\le y<2.$

For $0.02\le x<0.03,$ we want $$\lfloor y\rfloor \le 2,$$so $0\le y<3,$ and so on.

Thus, for $0\le x<1,$ the region is as follows.



The area of this part of the region is then $$0.01(1+2+3+\cdots +100)=0.01\cdot \frac{100\cdot 101}{2}.$$Next, we look at the interval where $1\le x<2,$ so $n=1.$ For $1\le x<1.01,$ we want $$\lfloor y\rfloor \le 0-1=-1,$$so there are no values of $y$ that work.

For $1.01\le x<1.02,$ we want $$\lfloor y\rfloor \le 1-1=0,$$so $0\le y<1.$

For $1.02\le x<1.03,$ we want $$\lfloor y\rfloor \le 2-1=1,$$so $0\le y<2,$ and so on.

Thus, for $1\le x<2,$ the region is as follows.



The area of this part of the region is then $$0.01(1+2+3+\cdots +99)=0.01\cdot \frac{99\cdot 100}{2}.$$Similarly, the area of the region for $2\le x<3$ is $$0.01(1+2+3+\cdots +98)=0.01\cdot \frac{98\cdot 99}{2},$$the area of the region for $3\le x<4$ is $$0.01(1+2+3+\cdots +97)=0.01\cdot \frac{97\cdot 98}{2},$$and so on, until the area of the region for $99\le x<100$ is $$0.01(1)=0.01\cdot \frac{1\cdot 2}{2}.$$Hence, the total area of the region is $$\frac{0.01}{2}(1\cdot 2+2\cdot 3+3\cdot 4+\cdots +100\cdot 101)=\frac{1}{200}\sum _{k=1}^{100}k(k+1).$$To compute this sum, we can use the formula $$\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}.$$Alternatively, we can write $$k(k+1)=\frac{(k+2)-(k-1)}{3}\cdot k(k+1)=\frac{k(k+1)(k+2)-(k-1)k(k+1)}{3},$$which allows sum to telescope, and we get $$\frac{1}{200}\sum _{k=1}^{100}k(k+1)=\frac{1}{200}\cdot \frac{100\cdot 101\cdot 102}{3}=\boxed{1717}.$$