A sequence consists of the digits $122333444455555\ldots$ such that the each positive integer $n$ is repeated $n$ times, in increasing order. Find the sum of the $4501$st and $4052$nd digits of this sequence.
1. **Identify the structure of the sequence:**
   The sequence is constructed such that each positive integer \( n \) is repeated \( n \) times. For example, the sequence starts as:
   \[
   1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots
   \]

2. **Determine the position of the last appearance of each number:**
   The position of the last appearance of the number \( n \) can be found by summing the first \( n \) natural numbers:
   \[
   \text{Position of last } n = \frac{n(n+1)}{2}
   \]
   For example, the last appearance of 9 is at:
   \[
   \frac{9 \cdot 10}{2} = 45
   \]

3. **Find the range of numbers that include the 4501st and 4052nd digits:**
   We need to find \( n \) such that:
   \[
   \frac{n(n+1)}{2} \approx 4501 \quad \text{and} \quad \frac{n(n+1)}{2} \approx 4052
   \]
   Since the numbers are large, we approximate \( \frac{n(n+1)}{2} \) as \( \frac{n^2}{2} \):
   \[
   \frac{n^2}{2} \approx 4501 \implies n^2 \approx 9002 \implies n \approx \sqrt{9002} \approx 95
   \]
   \[
   \frac{n^2}{2} \approx 4052 \implies n^2 \approx 8104 \implies n \approx \sqrt{8104} \approx 90
   \]

4. **Calculate the exact positions:**
   For \( n = 94 \):
   \[
   \frac{94 \cdot 95}{2} = 4465
   \]
   For \( n = 95 \):
   \[
   \frac{95 \cdot 96}{2} = 4560
   \]
   Thus, the 4501st digit is within the 95's.

   For \( n = 89 \):
   \[
   \frac{89 \cdot 90}{2} = 4005
   \]
   For \( n = 90 \):
   \[
   \frac{90 \cdot 91}{2} = 4095
   \]
   Thus, the 4052nd digit is within the 90's.

5. **Determine the exact digits:**
   - The 4501st digit is the 36th digit in the sequence of 95's (since \( 4501 - 4465 = 36 \)).
   - The 4052nd digit is the 47th digit in the sequence of 90's (since \( 4052 - 4005 = 47 \)).

6. **Sum the digits:**
   - The 4501st digit is 95.
   - The 4052nd digit is 90.
   \[
   9 + 0 = 9
   \]

The final answer is \(\boxed{9}\)