Note that $n$ contributes $n \cdot d(n)$ digits, where $d(n)$ is the number of digits of $n$. Then because $1+\cdots+99=4950$, we know that the digits of interest appear amongst copies of two digit numbers. Now for $10 \leq n \leq 99$, the number of digits in the subsequence up to the last copy of $n$ is $$1+2+3+\cdots+9+2 \cdot(10+\cdots+n)=2 \cdot(1+\cdots+n)-45=n^{2}+n-45$$ Since $67^{2}+67-45=4511$, the two digits are 6 and 7 in some order, so have sum 13.
A sequence consists of the digits $122333444455555 \ldots$ such that each positive integer $n$ is repeated $n$ times, in increasing order. Find the sum of the 4501st and 4052nd digits of this sequence.