Given the sum:
\[ \sum_{n=1}^N \left\lfloor \frac{1}{5} + \frac{n}{25} \right\rfloor = 25 \]
Simplify the expression within the floor function:
\[ \frac{1}{5} + \frac{n}{25} = \frac{5n + 1}{25} \]
The sum becomes:
\[ \sum_{n=1}^N \left\lfloor \frac{5n + 1}{25} \right\rfloor = 25 \]
For \( n = 1 \) to \( n = 19 \), the floor function evaluates to 0:
\[ \left\lfloor \frac{5n + 1}{25} \right\rfloor = 0 \quad \text{for } n \in [1, 19] \]
For \( n = 20 \) to \( n = 44 \), the floor function evaluates to 1:
\[ \left\lfloor \frac{5n + 1}{25} \right\rfloor = 1 \quad \text{for } n \in [20, 44] \]
Calculate the total sum:
\[ \sum_{n=1}^{N} \left\lfloor \frac{5n + 1}{25} \right\rfloor = \sum_{n=1}^{19} 0 + \sum_{n=20}^{44} 1 = 0 + (44 - 20 + 1) \times 1 = 0 + 25 = 25 \]
The value of \( N \) that satisfies the equation is:
\[ \boxed{N = 44} \]