Let $n \geq 2, n \in \mathbb{N}$, $a,b,c,d \in \mathbb{N}$, $\frac{a}{b} + \frac{c}{d} < 1$ and $a + c \leq n,$ find the maximum value of $\frac{a}{b} + \frac{c}{d}$ for fixed $n.$
To find the maximum value of \(\frac{a}{b} + \frac{c}{d}\) for fixed \(n\) given the constraints \(n \geq 2\), \(a, b, c, d \in \mathbb{N}\), \(\frac{a}{b} + \frac{c}{d} < 1\), and \(a + c \leq n\), we proceed as follows:

1. **Substitute \(b\) and \(d\) with expressions involving \(a\) and \(c\):**
   Let \(b = a + x\) and \(d = c + y\). This transforms the inequality into:
   \[
   \frac{a}{a + x} + \frac{c}{c + y} < 1
   \]
   which simplifies to:
   \[
   \frac{a}{a + x} + \frac{c}{c + y} < 1 \iff xy > ac
   \]

2. **Assume \(a \leq c\):**
   We will show that the form \(\frac{a}{a + x} + \frac{c}{c + y}\) is maximal for \(x = \pi\) and \(y = 1\), where \(\pi = xy\).

3. **Maximize the expression:**
   Consider the function:
   \[
   f(x) = \frac{a}{a + x} + \frac{c}{c + y}
   \]
   We need to show that this function is maximized when \(x = \pi\) and \(y = 1\). This is equivalent to showing:
   \[
   \frac{a}{a + \pi} + \frac{c}{c + 1} \geq \frac{a}{a + x} + \frac{c}{c + y}
   \]
   which simplifies to:
   \[
   ay + cx \leq a + c\pi
   \]

4. **Use convexity:**
   Using the convex function \(f(x) = cx + \frac{a\pi}{x}\), we see that the maximum on \([1, \pi]\) is attained at one of the extremities, specifically when \(x = \pi\).

5. **Substitute \(\pi = ac + 1\):**
   Substituting \(\pi = ac + 1\) into the expression, we get:
   \[
   \frac{a}{a + \pi} + \frac{c}{c + 1} = 1 - \frac{1}{t}
   \]
   where \(t = (ac + a + 1)(c + 1)\).

6. **Maximize \(t\):**
   To maximize \(t\), we note that it is increasing in \(a\) and \(c\). Therefore, we may assume \(a + c = n\).

7. **Simplify using new variables:**
   Let \(\gamma = c + 1\) and \(\sigma = n + 1\). We then find:
   \[
   t = \gamma(a\gamma + 1)
   \]
   Consider the function \(g(x) = x + x^2(\sigma - x)\) on \([ \frac{\sigma}{2}, \sigma ]\).

8. **Find the maximum of \(g(x)\):**
   The function \(g(x)\) is concave on this interval with a maximum at:
   \[
   x = \frac{\sigma + \sqrt{\sigma^2 + 3}}{3}
   \]
   Notice that:
   \[
   \frac{2\sigma}{3} < x < \frac{2\sigma + 1}{3}
   \]
   Since we consider integers, we split up with regards to \(n\) modulo 3.

9. **Evaluate for different cases of \(n\) modulo 3:**
   - If \(n = 3p + 1\), then \(\sigma = 3p + 2\) and \(\frac{2p + 1}{3} < x < \frac{2p + 2}{3}\). We need to check \(x = 2p + 1\) and \(x = 2p + 2\) to see which yields the bigger value.
   - Similarly, evaluate for other cases modulo 3.

10. **Calculate \(t\) for each case:**
    \[
    t(3p) = 4p^3 + 4p^2 + 3p + 1
    \]
    \[
    t(3p + 1) = 4p^3 + 8p^2 + 7p + 2
    \]
    \[
    t(3p + 2) = 4p^3 + 12p^2 + 14p + 6
    \]

The final answer is \(\boxed{\frac{a}{a + \pi} + \frac{c}{c + 1}}\) where \(\pi = ac + 1\).