For a given positive integer $n,$ find 
$$\sum_{k=0}^{n} \left(\frac{\binom{n}{k} \cdot (-1)^k}{(n+1-k)^2} - \frac{(-1)^n}{(k+1)(n+1)}\right).$$
To solve the given sum, we start by defining two functions \( f(x) \) and \( g(x) \) and then use these functions to simplify the given expression.

1. Define the function \( f(x) \) as follows:
   \[
   f(x) = \sum_{k=0}^n \binom{n}{k} \frac{x^{k+1}}{(k+1)^2}
   \]

2. Note that:
   \[
   (xf'(x))' = \sum_{k=0}^n \binom{n}{k} x^k = (1+x)^n
   \]

3. Integrate both sides with respect to \( x \):
   \[
   xf'(x) = \int_{0}^x (1+t)^n \, dt = \frac{(1+x)^{n+1} - 1}{n+1}
   \]

4. Integrate again to find \( f(x) \):
   \[
   f(x) = \int_{0}^x \frac{(1+t)^{n+1} - 1}{(n+1)t} \, dt
   \]

5. Define another function \( g(x) \) as follows:
   \[
   g(x) = \sum_{k=0}^n \frac{x^{k+1}}{k+1}
   \]

6. Differentiate \( g(x) \):
   \[
   g'(x) = \sum_{k=0}^n x^k = \frac{x^{n+1} - 1}{x - 1}
   \]

7. Integrate \( g'(x) \) to find \( g(x) \):
   \[
   g(x) = \int_{0}^x \frac{t^{n+1} - 1}{t - 1} \, dt
   \]

8. Now, consider the given sum \( S \):
   \[
   S = \sum_{k=0}^n \left( \frac{\binom{n}{k} \cdot (-1)^k}{(n+1-k)^2} - \frac{(-1)^n}{(k+1)(n+1)} \right)
   \]

9. Rewrite \( S \) using \( f(x) \) and \( g(x) \):
   \[
   S = (-1)^{n+1} f(-1) + \frac{(-1)^{n+1}}{n+1} g(1)
   \]

10. Evaluate \( f(-1) \):
    \[
    f(-1) = \int_{0}^{-1} \frac{(1+t)^{n+1} - 1}{(n+1)t} \, dt
    \]

11. Change the variable of integration \( t \) to \( u = -t \):
    \[
    f(-1) = \int_{0}^{1} \frac{(1-u)^{n+1} - 1}{(n+1)(-u)} \, du = -\int_{0}^{1} \frac{(1-u)^{n+1} - 1}{(n+1)u} \, du
    \]

12. Evaluate \( g(1) \):
    \[
    g(1) = \int_{0}^{1} \frac{t^{n+1} - 1}{t - 1} \, dt
    \]

13. Combine the results:
    \[
    S = (-1)^{n+1} \left( -\int_{0}^{1} \frac{(1-u)^{n+1} - 1}{(n+1)u} \, du \right) + \frac{(-1)^{n+1}}{n+1} \int_{0}^{1} \frac{t^{n+1} - 1}{t - 1} \, dt
    \]

14. Simplify the expression:
    \[
    S = \frac{(-1)^n}{n+1} \int_{0}^{1} \frac{u^n - 1}{u - 1} \, du + \frac{(-1)^{n+1}}{n+1} \int_{0}^{1} \frac{t^n - 1}{t - 1} \, dt = 0
    \]

The final answer is \(\boxed{0}\).