Decide whether $S_n$ or $T_n$ is larger, where
\[S_n =\displaystyle\sum_{k=1}^n \frac{k}{(2n - 2k + 1)(2n - k + 1)}, T_n =\displaystyle\sum_{k=1}^n\frac{1}{k}\]
1. We start with the given expressions for \( S_n \) and \( T_n \):
   \[
   S_n = \sum_{k=1}^n \frac{k}{(2n - 2k + 1)(2n - k + 1)}, \quad T_n = \sum_{k=1}^n \frac{1}{k}
   \]

2. We need to compare \( S_n \) and \( T_n \). First, let's simplify \( S_n \):
   \[
   S_n = \sum_{k=1}^n \frac{k}{(2n - 2k + 1)(2n - k + 1)}
   \]

3. Notice that the term inside the summation for \( S_n \) can be decomposed using partial fractions:
   \[
   \frac{k}{(2n - 2k + 1)(2n - k + 1)} = \frac{A}{2n - 2k + 1} + \frac{B}{2n - k + 1}
   \]
   Solving for \( A \) and \( B \), we get:
   \[
   k = A(2n - k + 1) + B(2n - 2k + 1)
   \]
   By equating coefficients, we find:
   \[
   A = 1, \quad B = -1
   \]
   Therefore:
   \[
   \frac{k}{(2n - 2k + 1)(2n - k + 1)} = \frac{1}{2n - 2k + 1} - \frac{1}{2n - k + 1}
   \]

4. Substituting back into the summation for \( S_n \):
   \[
   S_n = \sum_{k=1}^n \left( \frac{1}{2n - 2k + 1} - \frac{1}{2n - k + 1} \right)
   \]

5. This can be written as a telescoping series:
   \[
   S_n = \left( \frac{1}{2n - 1} - \frac{1}{2n} \right) + \left( \frac{1}{2n - 3} - \frac{1}{2n - 1} \right) + \cdots + \left( \frac{1}{1} - \frac{1}{n + 1} \right)
   \]

6. The terms in the series cancel out, leaving:
   \[
   S_n = \sum_{k=1}^n \frac{1}{2k - 1} - \sum_{k=1}^n \frac{1}{2n - k + 1}
   \]

7. Recognize that the second sum can be rewritten:
   \[
   \sum_{k=1}^n \frac{1}{2n - k + 1} = \sum_{k=n+1}^{2n} \frac{1}{k}
   \]

8. Therefore:
   \[
   S_n = \sum_{k=1}^n \frac{1}{2k - 1} - \sum_{k=n+1}^{2n} \frac{1}{k}
   \]

9. Using the harmonic series properties, we can write:
   \[
   \sum_{k=1}^{2n} \frac{1}{k} = \sum_{k=1}^n \frac{1}{k} + \sum_{k=n+1}^{2n} \frac{1}{k}
   \]

10. Thus:
    \[
    S_n = \sum_{k=1}^{2n} \frac{1}{k} - \sum_{k=n+1}^{2n} \frac{1}{k} - \sum_{k=1}^n \frac{1}{2k}
    \]

11. Simplifying further:
    \[
    S_n = \sum_{k=1}^n \frac{1}{k} - \sum_{k=1}^n \frac{1}{2k}
    \]

12. Finally, we compare \( S_n \) and \( T_n \):
    \[
    T_n = \sum_{k=1}^n \frac{1}{k}
    \]
    \[
    S_n = \sum_{k=1}^n \frac{1}{k} - \sum_{k=1}^n \frac{1}{2k}
    \]

13. Clearly:
    \[
    S_n < T_n
    \]

Conclusion:
\[
\boxed{T_n > S_n}
\]