Suppose that $n$ is a positive integer and that $a$ is the integer equal to $\frac{10^{2n}-1}{3\left(10^n+1\right)}.$
If the sum of the digits of $a$ is 567, what is the value of $n$?
1. We start with the given expression for \( a \):
   \[
   a = \frac{10^{2n} - 1}{3(10^n + 1)}
   \]

2. We simplify the numerator \( 10^{2n} - 1 \) using the difference of squares:
   \[
   10^{2n} - 1 = (10^n)^2 - 1^2 = (10^n - 1)(10^n + 1)
   \]

3. Substituting this back into the expression for \( a \):
   \[
   a = \frac{(10^n - 1)(10^n + 1)}{3(10^n + 1)}
   \]

4. Since \( 10^n + 1 \) is a common factor in both the numerator and the denominator, we can cancel it out:
   \[
   a = \frac{10^n - 1}{3}
   \]

5. Next, we analyze the form of \( 10^n - 1 \). This is a number consisting of \( n \) digits of 9. For example, if \( n = 3 \), then \( 10^3 - 1 = 999 \).

6. Dividing \( 10^n - 1 \) by 3, we get a number consisting of \( n \) digits of 3. For example, if \( n = 3 \), then \( \frac{999}{3} = 333 \).

7. We are given that the sum of the digits of \( a \) is 567. Since \( a \) consists of \( n \) digits of 3, the sum of the digits of \( a \) is:
   \[
   3n = 567
   \]

8. Solving for \( n \):
   \[
   n = \frac{567}{3} = 189
   \]

The final answer is \( \boxed{189} \)