Just saw this post, and realized that

1/9801 =
0.(000102030405060708091011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768697071727374757677787980818283848586878889909192939495969799)(repeat)

Similar properties are also exhibited by numbers 998001, 99980001, .. and so on.
Edit: fixed periodic start and end
It is not very obvious to me why this happens. Is there some simple explanation to this property?

There is actually a straightforward reason.   As $99$ is $1$ less than $100$, we get a fairly simple expression for its decimal $$\frac{1}{99}=0.01010101010101\overline{01}\dots$$  Now, $$\frac{1}{9801}=\left(\frac{1}{99}\right)^2,$$ and the decimal expansion follows from the formula for general power series $$\left(\sum_{n=1}^\infty x^n\right)^2= x\sum_{n=1}^\infty nx^n.$$ 

Letting $x=\frac{1}{100}$, the decimal expansion for $\frac{1}{99}$ given above is exactly the same thing as writing $\frac{1}{99}=\sum_{n=1}^\infty x^n$.  Applying our identity, the $x$ in front accounts for the double zero.  Once $n$ is around $99$ we expect to miss a number because we are forcing things to be in decimal, and there will be carrying, which is why the number 98 is missed.

A similar pattern will occur for $\frac{1}{998001}=\left(\frac{1}{999}\right)^2,$ since as before $$\frac{1}{999}=0.001001\overline{001}.$$
