For any $n,T \geq 2, n, T \in \mathbb{N}$, find all $a \in \mathbb{N}$ such that $\forall a_i > 0, i = 1, 2, \ldots, n$, we have

\[\sum^n_{k=1} \frac{a \cdot k + \frac{a^2}{4}}{S_k} < T^2 \cdot \sum^n_{k=1} \frac{1}{a_k},\] where $S_k = \sum^k_{i=1} a_i.$
1. We start with the given inequality:
   \[
   \sum_{k=1}^n \frac{a \cdot k + \frac{a^2}{4}}{S_k} < T^2 \cdot \sum_{k=1}^n \frac{1}{a_k},
   \]
   where \( S_k = \sum_{i=1}^k a_i \).

2. We know that \( T^2 \geq 4 \) since \( T \geq 2 \). Therefore, we can rewrite the inequality as:
   \[
   \sum_{k=1}^n \frac{a \cdot k + \frac{a^2}{4}}{S_k} < 4 \cdot \sum_{k=1}^n \frac{1}{a_k}.
   \]

3. We decompose the left-hand side:
   \[
   \sum_{k=1}^n \frac{a \cdot k + \frac{a^2}{4}}{S_k} = \sum_{k=1}^n \left( \frac{a \cdot k}{S_k} + \frac{a^2}{4S_k} \right).
   \]

4. This can be split into two sums:
   \[
   \sum_{k=1}^n \frac{a \cdot k}{S_k} + \sum_{k=1}^n \frac{a^2}{4S_k}.
   \]

5. We need to show that:
   \[
   \sum_{k=1}^n \frac{a \cdot k}{S_k} + \sum_{k=1}^n \frac{a^2}{4S_k} < 4 \cdot \sum_{k=1}^n \frac{1}{a_k}.
   \]

6. Consider the first sum:
   \[
   \sum_{k=1}^n \frac{a \cdot k}{S_k}.
   \]
   Since \( S_k = \sum_{i=1}^k a_i \), we know that \( S_k \geq a_k \). Therefore:
   \[
   \frac{k}{S_k} \leq \frac{k}{a_k}.
   \]

7. Summing over \( k \) from 1 to \( n \):
   \[
   \sum_{k=1}^n \frac{k}{S_k} \leq \sum_{k=1}^n \frac{k}{a_k}.
   \]

8. Now consider the second sum:
   \[
   \sum_{k=1}^n \frac{a^2}{4S_k}.
   \]
   Using the same logic, since \( S_k \geq a_k \):
   \[
   \frac{1}{S_k} \leq \frac{1}{a_k}.
   \]

9. Summing over \( k \) from 1 to \( n \):
   \[
   \sum_{k=1}^n \frac{1}{S_k} \leq \sum_{k=1}^n \frac{1}{a_k}.
   \]

10. Combining these results, we get:
    \[
    \sum_{k=1}^n \frac{a \cdot k}{S_k} + \sum_{k=1}^n \frac{a^2}{4S_k} \leq a \sum_{k=1}^n \frac{k}{a_k} + \frac{a^2}{4} \sum_{k=1}^n \frac{1}{a_k}.
    \]

11. We need to show that this is less than \( 4 \sum_{k=1}^n \frac{1}{a_k} \):
    \[
    a \sum_{k=1}^n \frac{k}{a_k} + \frac{a^2}{4} \sum_{k=1}^n \frac{1}{a_k} < 4 \sum_{k=1}^n \frac{1}{a_k}.
    \]

12. Simplifying, we get:
    \[
    a \sum_{k=1}^n \frac{k}{a_k} + \frac{a^2}{4} \sum_{k=1}^n \frac{1}{a_k} < 4 \sum_{k=1}^n \frac{1}{a_k}.
    \]

13. Dividing both sides by \( \sum_{k=1}^n \frac{1}{a_k} \):
    \[
    a \sum_{k=1}^n \frac{k}{a_k} \cdot \frac{1}{\sum_{k=1}^n \frac{1}{a_k}} + \frac{a^2}{4} < 4.
    \]

14. Since \( \sum_{k=1}^n \frac{k}{a_k} \cdot \frac{1}{\sum_{k=1}^n \frac{1}{a_k}} \leq n \):
    \[
    a \cdot n + \frac{a^2}{4} < 4.
    \]

15. Solving for \( a \):
    \[
    a \cdot n + \frac{a^2}{4} < 4.
    \]

16. This inequality holds for \( a = 1 \) and \( a = 2 \) when \( n \leq 3 \).

The final answer is \( \boxed{ a = 1 } \) and \( a = 2 \).