It is easy to observe that $\delta(n)$ is always less than $10$. If you partition the digits in two groups with difference of sums $\delta(n)$ and $\delta(n)\geq 10$ then move one of the digits from the group with bigger sum to the other group. Hence $\delta(n)$ is bounded by $10$. I claim that the average is $\frac{1}{2}$. It can be seen that this number is bigger than or equal to $\frac{1}{2}$, because if the sum of the digits of $n$ is odd then $\delta(n)\geq 1$, and the probability of having an odd digital sum is $\frac{1}{2}$. One can also see that the probability of a number to have at least $10$ digits of $1$, is $1$ (Really easy to show). Now the average of $\delta(n)$ over all numbers will be equal to the average over the special numbers I mentioned (The ones having at least ten digits of $1$) For each of these numbers like $n$, just remove the first ten digits of $1$ to obtain $m$. Then we know that $\delta(m)<10$. Now using the extra $1$ digits we have at hand we can distribute them among the two groups we have for $\delta(m)$ to decrease the difference between sums to either $0$ or $1$. If the sum of digits of $m$ is odd we can reduce $\delta(m)$ to $1$ and in the other case to $0$. One can again easily observe that the probability of $m$ having an odd digital sum is $\frac{1}{2}$. So for half of the numbers (with respect to probability) we have $\delta(n)=0$ and for another half (disjoint from the last half) we have $\delta(n)\leq 1$. So the average we are trying to find is less than or equal to $\frac{1}{2}$ and hence my claim is proved.

I find $ \displaystyle\lim_{N \rightarrow \infty}\displaystyle\frac{\displaystyle\sum_{k = 1}^{10^N}\delta(k)}{10^N} $. Notice that $\forall k$, $\delta(k) \le 9$, since if it is more than $9$, and the digits of $k$ are partitioned into $X \cup Y$, then we can let $X' = X \setminus \{a\}, Y'=Y \setminus \{a\}$ for any $a \in X$, and the defect will be smaller. Notice that, out of the $10^N$ numbers from $1$ to $10^N$, at most $9\cdot 9^{N}$ of them don't have all the digits. For these numbers, I use the bound $\delta(k) \le 9$. For any number $k$ that does have all the digits, I prove $\delta(k) \le 1$, by contradiction, assuming $\delta(k) \ge 2$. Let $X \cup Y$ be an optimal partition, with $\sum X - \sum Y = \delta(k) \ge 2$. If $1 \in X$, then we can put 1 in $Y$ instead of $X$, thus decreasing the defect by $1$. Since the defect was at least 2, I have strictly decreased the defect, contradiction. But since $1$ appears in $k$, then $1 \in Y$. If $2 \in X$, I can interchange 1 and 2 in $X$ and $Y$, and the defect will decrease by 1, contradiction. But since $2$ appears in $k$, then $2 \in Y$. I can continue and I will get that $1,2,...,9$ don't appear in $X$ but they do appear in $Y$. From this, I get that $\sum X=0, \sum Y >0$, and thus $\delta(k) = \sum X - \sum Y $ is impossible! Thus I have proven $\delta(k) \le 1$. Also, if the sum of digits of $k$ is odd, then clearly $\delta(k)=1$ and if it is even, then $\delta(k)=0$, by parity. Notice that half of the numbers satisfy the first condition and half satisfy the second condition. Therefore, $\displaystyle\frac{\displaystyle\sum_{k = 1}^{10^N}\delta(k)}{10^N} \le \left( \displaystyle\frac{1}{10^N} \right) \cdot \left( 9*(9^{N}) + 0*\frac{10^N-9^N}{2}+1*\frac{10^N-9^N}{2} \right) $, and, since $\delta(k) \ge 1$ if the sum of digits of $k$ is odd, which occurs for half of the numbers, $\displaystyle\frac{\displaystyle\sum_{k = 1}^{10^N}\delta(k)}{10^N} \ge \left( \displaystyle\frac{1}{10^N} \right) \cdot \left(1*\frac{10^N}{2} \right) $, both of which tend to $1/2$, so we're done.