Given an integer $N \geq 4$, determine the largest value the sum
$$\sum_{i=1}^{\left \lfloor{\frac{k}{2}}\right \rfloor+1}\left( \left \lfloor{\frac{n_i}{2}}\right \rfloor+1\right)$$
may achieve, where $k, n_1, \ldots, n_k$ run through the integers subject to $k \geq 3$, $n_1 \geq \ldots\geq n_k\geq 1$ and $n_1 + \ldots + n_k = N$.
To determine the largest value of the sum 
\[ \sum_{i=1}^{\left \lfloor{\frac{k}{2}}\right \rfloor+1}\left( \left \lfloor{\frac{n_i}{2}}\right \rfloor+1\right) \]
where \( k, n_1, \ldots, n_k \) run through the integers subject to \( k \geq 3 \), \( n_1 \geq \ldots \geq n_k \geq 1 \), and \( n_1 + \ldots + n_k = N \), we proceed as follows:

1. **Assume \( k \) is even**: Let \( k = 2m \). This assumption simplifies the problem by reducing the number of cases we need to consider.

2. **Assume \( n_i = 1 \) for \( i > m+1 \)**: This assumption is based on the idea that having smaller values for \( n_i \) beyond a certain point will not contribute significantly to the sum.

3. **Assume \( n_{m+1} > 1 \)**: This ensures that the sum is maximized by having at least one larger value among the \( n_i \).

4. **Calculate the sum**: Given the assumptions, we need to calculate the sum for the first \( m+1 \) terms:
   \[
   \sum_{i=1}^{m+1} \left( \left \lfloor{\frac{n_i}{2}}\right \rfloor + 1 \right)
   \]
   Since \( n_1 + \ldots + n_k = N \), we need to distribute \( N \) among the \( n_i \) values.

5. **Maximize the sum**: To maximize the sum, we need to find the optimal distribution of \( N \) among the \( n_i \) values. We consider the cases based on the residue of \( N \) modulo 3.

   - **Case 1: \( N = 3m + 1 \)**:
     \[
     \sum_{i=1}^{m+1} \left( \left \lfloor{\frac{n_i}{2}}\right \rfloor + 1 \right) = 2m + 2
     \]
   - **Case 2: \( N = 3m + 2 \)**:
     \[
     \sum_{i=1}^{m+1} \left( \left \lfloor{\frac{n_i}{2}}\right \rfloor + 1 \right) = 2m + 2
     \]
   - **Case 3: \( N = 3m + 3 \)**:
     \[
     \sum_{i=1}^{m+1} \left( \left \lfloor{\frac{n_i}{2}}\right \rfloor + 1 \right) = 2m + 3
     \]

6. **Conclusion**: The largest value the sum can achieve is given by:
   \[
   f(3m+1) = 2m + 2, \quad f(3m+2) = 2m + 2, \quad f(3m+3) = 2m + 3
   \]

The final answer is \( \boxed{ f(3m+1) = 2m + 2, f(3m+2) = 2m + 2, f(3m+3) = 2m + 3 } \)