What is the relationship between 1/7, 1/11, 1/13, and the number 1001?
The relationship is that $\frac17=\frac{11\times13}{1001}$, $\frac1{11}=\frac{7\times13}{1001}$ and $\frac1{13}=\frac{7\times11}{1001}$, and the pattern of adding three digits to the next three digits to get $999$ works for any fraction $\frac{n}{1001}$ (apart from the ones that are integers, like $\frac{1001}{1001}$ or $\frac{2002}{1001}$) and for no other number.
Let's see what multiplying any number by $1001$ does. We have $$ 1001x=(1000+1)x=1000x+x $$ So multiplying a number $x$ by $1001$ is the same as taking two copies of $x$, multiply one of them by $1000$ (which has the effect of moving all the digits of $x$ three places to the left), then add them together.
Now, what's special about the numbers $\frac n{1001}$ (again, apart from the ones which are already integers) is that they are the only numbers which are not integers, but multiplying them by $1001$ makes them integers. If we use the above interpretation of multiplying by $1001$, the only way that can happen is if after doing the addition, we're left with something with decimal part $.999999\ldots$ (you can try to find examples which make it $.000000\ldots$, but you will not succeed). This means exactly that any three digits of the decimal expansion of $\frac n{1001}$, plus the next three digits must add up to $999$.
(They could, conceivably, add up to $1998$, and let the $1$ carry into the next set of three, but since $1998=999+999$, that means that we would have to start with $.999999\ldots$ already, meaning what we have is an integer. And we're not considering those.)
Knowing how this phenomenon came to happen (e.g. as explained in Arthurs answer), you can construct similar ones. E.g. You can factor
$$10^\color{red}4+1=73\times 137.$$
You will observe that subdividing the decimal digits of the reciprocals of these factors into blocks of length $\color{red}{\text{four}}$ will give you sums of $9,\!999$:
\begin{align} 1/73& = 0.\underline{0136}\overline{9863}\underline{0136}\overline{9863}...\\[1ex] 1/137 &= 0.\underline{0072}\overline{9927}\underline{0072}\overline{9927}... \end{align}
Further, $1/7$, $1/11$ and $1/13$ also are special for other block lengths: e.g. they give the sum $999,\!999,\!999$ when using block length $\color{blue}{\text{nine}}$ (not very surprising). However, there are other such numbers: $1/19$ and $1/52579$. The reason is that
$$10^\color{blue}9+1=7\times 11\times 13\times 19\times 52579.$$