These are some odd digits

Let $f(999~999~991,~n)=x$ .

We know intuitively that $f(999~999~991,~n)$ must be an odd number.

Then, we need to do a little bit of trick.

$\begin{aligned} 999~999~991x &=1~000~000~000x-9x \\ & = 1~000~000~000(x-1)+1~000~000~000-9x \end{aligned}$

Suppose that $x\le 111~111~111.$

$0<1~000~000~000-9x<1~000~000~000$ and therefore $1~000~000~000-9x$ cannot have any effect on the digits over billion.

And since $x-1$ is an even number, the billionth digit must be an even number. This doesn't make sense, therefore $x\ge 111~111~113.$

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Then let's assume that $111~111~113\le x \le 222~222~222.$

Note that

$1\times9={\color{#D61F06}{0}}{\color{#3D99F6}{9}}; \\ 3\times9={\color{#D61F06}{2}}{\color{#3D99F6}{7}}; \\ 5\times9={\color{#D61F06}{4}}{\color{#3D99F6}{5}}; \\ 7\times9={\color{#D61F06}{6}}{\color{#3D99F6}{3}}; \\ 9\times9={\color{#D61F06}{8}}{\color{#3D99F6}{1}}.$

And therefore if the last two digits of $x$ is odd, the last two digits of $9x$ is also odd.

But then note that (yes there're a lot of things to note, sorry.)

$\boxed{\text{Note A.}} \\ \\ \begin{aligned} 999~999~991x &=1~000~000~000x-9x \\ & = 1~000~000~000(x-2) + 2~000~000~000 - 9x \end{aligned}$

So the last $9$ digits of $999~999~991x$ is the same as $2~000~000~000-9x$ , which is larger than $0$ but also smaller than $1~000~000~000$ . Then again, no effects on the digits over billionth.

Also, if the last two digits of $9x$ is odd, the tenth digit of $2~000~000~000-9x$ must be even.

Therefore, the tenth digit of $x$ cannot be odd. (since the unit's digit is always odd.)

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Then by the information we know, the smallest possible that $x$ can be is $111~111~121.$

But then by <Note A>, we also know that ${\color{#D61F06}{x-2~\text{should only have odd digits.}}}$

Therefore we can say that $x$ follows the form ${\color{#D61F06}{\overline{*******[\text{even number}]1}}}.$

Possible values for $x$ are: $\boxed{111~111~121,~111~111~141,~111~111~161,~111~111~181}.$

We can also see that $\boxed{111~111~201}$ maybe would satisfy the condition.

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What about $111~111~321$ then?

Sadly it can't go into the answers' team.

Why, you may ask.

Calculate $1321\times 9$ and you get $1\boxed{1}889$ , and that $\boxed{1}$ would leave the thousandth digit of $9x$ as an odd number, which shouldn't happen, because then $2 000 000 000-9x$ would have its thousandth digit as an even number.

But from $1341\times 9$ , things get better again and we get $1\boxed{2}069.$

Then possible values for $x$ in this section are: $\boxed{111~111~341,~111~111~361,~111~111~381,~111~111~401}.$

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Do the same with above and we see that $111~111~521$ and $111~111~541$ are impossible.

Then possible values for $x$ in this section are: $\boxed{111~111~561,~111~111~581,~111~111~601}.$

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From above, I only guessed, using the terms "possible values" and "maybe would satisfy".

But they really do satisfy the conditions. Reflect on and enjoy these beautiful numbers.

$\begin{aligned} 999~999~991\times111~111~121 &= 111~111~119~999~999~911 \\ 999~999~991\times111~111~141 &= 111~111~139~999~999~731 \\ 999~999~991\times111~111~161 &= 111~111~159~999~999~551 \\ 999~999~991\times111~111~181 &= 111~111~179~999~999~371 \\ 999~999~991\times111~111~201 &= 111~111~199~999~999~191 \\ \\ 999~999~991\times111~111~341 &= 111~111~339~999~997~931 \\ 999~999~991\times111~111~361 &= 111~111~359~999~997~751 \\ 999~999~991\times111~111~381 &= 111~111~379~999~997~571 \\ 999~999~991\times111~111~401 &= 111~111~399~999~997~391 \\ \\ 999~999~991\times111~111~561 &= 111~111~559~999~995~951 \\ 999~999~991\times111~111~581 &= 111~111~579~999~995~771 \\ 999~999~991\times111~111~601 &= 111~111~599~999~995~591 \end{aligned}$

See the pattern? It's so beautiful! ...to my eyes, at least.

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So anyway, the answer to this problem is:

$\begin{aligned} \sum_{n=1}^{10} f(999~999~991,~n) &=111~111~000\times10+121+141+161+181+201+341+361+381+401+561 \\ &=\boxed{1~111~112~850}. \end{aligned}$