Suppose an integer \(n\) has \(k\) digits and let \(a_k\) be the \(k\)th digit. We can represent \(n\) like this:
$$\begin{align} n &= a_1 + a_2 10^1 + a_3 10^2 + a_4 10^3 + a_5 10^4 + a_6 10^5 + \cdots \\ n &= (a_1 + a_3 10^2 + a_5 10^4 + a_7 10^6 + \cdots) + (a_2 10 + a_4 10^3 + a_6 10 ^5 + \cdots) \end{align}$$
Before going any further, I am going to make two claims.
Claim 1: An integer made of even number of 9's is divisible by 11:
$$\begin{align} 11 &| 99 \\ 11 &| 9999 &&\text{ because } 11 | 99 \implies 11| 9900 \implies 11 | (9900 + 99) \\ 11 &| 999999 &&\text{ because } 11 | 9999 \implies 11| 999900 \implies 11 | (999900 + 99) \end{align}$$
Claim 2: An integer with even digits in the form of \(10\ldots01\) (like 1001 or 100001) is divisible by 11:
$$\begin{align} 11 &| 99 &&\implies 11 |990 && \implies 11 |990+11 && \implies 11 |1001 \\ 11 & | 9999 && \implies 11 |99990 && \implies 11 |99990+11 && \implies 11 |100001\\ 11 & | 999999 && \implies 11 | 9999990 && \implies 11 |9999990+11 && \implies 11 |10000001 \end{align}$$
Now let's divide \(n\) by 11:
$$\frac{n}{11} = \frac{(a_1 + a_3 10^2 + a_5 10^4 + a_7 10^6 + \cdots)}{11} + \frac{(a_2 10 + a_4 10^3 + a_6 10 ^5 + \cdots)}{11}$$
Since \((10^k = 99\ldots +1)\), where \((99\ldots)\) has \(k\) digits:
$$\frac{n}{11} = \frac{(a_3 99 + a_5 9999 + a_7 999999 + \cdots)}{11} + \frac{(a_1 + a_3 + a_5 + \cdots)}{11} + \frac{(a_2 10 + a_4 10^3 + a_6 10 ^5 + \cdots)}{11}$$
Using claim 1, there exists an integer \(h\) such that:
$$\frac{n}{11} = h + \frac{(a_1 + a_3 + a_5 + \cdots)}{11} + \frac{(a_2 10 + a_4 10^3 + a_6 10^5 + \cdots)}{11}$$
Rearranging:
$$\begin{gather} \frac{n}{11} = h + \frac{(a_1 + a_3 + a_5 + \cdots)}{11} + \frac{(a_2 10 + a_4 1000 + a_6 100000 + \cdots)}{11} \\ \frac{n}{11} = h + \frac{(a_1 + a_3 + a_5 + \cdots)}{11} + \frac{(a_2 11 + a_4 1001 + a_6 100001 + \cdots)}{11} - \frac{(a_2 + a_4 + a_6 + \cdots)}{11} \end{gather}$$
Using claim 2, there exists an integer \(j\) such that:
$$\begin{gather} \frac{n}{11} = h + \frac{(a_1 + a_3 + a_5 + \cdots)}{11} + j - \frac{(a_2 + a_4 + a_6 + \cdots)}{11} \\ \frac{n}{11} = h + j + \frac{(a_1 - a_2 + a_3 - a_4 + a_5 - a_6 + \cdots)}{11} \end{gather}$$
This means \(\frac{n}{11}\) would an integer if and only if \(\frac{(a_1 - a_2 + a_3 - a_4 + a_5 - a_6 + \cdots)}{11}\) is an integer.