Find all natural $a,b$ such that $\left. {a(a + b) + 1} \right|(a + b)(b + 1) - 1$.
1. We start by rewriting the given condition: 
   \[
   a(a + b) + 1 \mid (a + b)(b + 1) - 1
   \]
   This means that \(a(a + b) + 1\) divides \((a + b)(b + 1) - 1\).

2. Let us express \(b\) in terms of \(a\). Write \(b = aq + r\), where \(q, r \in \mathbb{N} \cup \{0\}\) and \(0 \leq r < a\). This implies that \(b \geq aq\).

3. Substitute \(b = aq + r\) into the expression \((a + b)(b + 1) - 1\):
   \[
   (a + b)(b + 1) - 1 = (a + aq + r)(aq + r + 1) - 1
   \]

4. Simplify the expression:
   \[
   (a + aq + r)(aq + r + 1) - 1 = (a(1 + q) + r)(aq + r + 1) - 1
   \]
   \[
   = (a + aq + r)(aq + r + 1) - 1
   \]
   \[
   = a(aq + r + 1) + aq(aq + r + 1) + r(aq + r + 1) - 1
   \]
   \[
   = a^2q + ar + a + a^2q^2 + aqr + aq + arq + r^2 + r - 1
   \]

5. Now, consider the divisor \(a(a + b) + 1\):
   \[
   a(a + b) + 1 = a(a + aq + r) + 1 = a^2 + a^2q + ar + 1
   \]

6. We need to check if \(a(a + b) + 1\) divides \((a + b)(b + 1) - 1\). We can write:
   \[
   (a + b)(b + 1) - 1 = k(a(a + b) + 1)
   \]
   for some integer \(k\).

7. Substitute \(b = aq + r\) into the equation:
   \[
   (a + aq + r)(aq + r + 1) - 1 = k(a^2 + a^2q + ar + 1)
   \]

8. Simplify the left-hand side:
   \[
   a^2q + ar + a + a^2q^2 + aqr + aq + arq + r^2 + r - 1 = k(a^2 + a^2q + ar + 1)
   \]

9. Equate the coefficients of the terms on both sides. This will give us a system of equations to solve for \(a\) and \(b\).

10. From the problem, we see that if \(a = 1\) and \(q = b\), the equation simplifies significantly. Substitute \(a = 1\):
    \[
    (1 + b)(b + 1) - 1 = k(1(1 + b) + 1)
    \]
    \[
    (1 + b)(b + 1) - 1 = k(1 + b + 1)
    \]
    \[
    b^2 + 2b + 1 - 1 = k(b + 2)
    \]
    \[
    b^2 + 2b = k(b + 2)
    \]

11. Solve for \(k\):
    \[
    b^2 + 2b = kb + 2k
    \]
    \[
    b^2 + 2b - kb - 2k = 0
    \]
    \[
    b(b + 2 - k) = 2k
    \]

12. For \(b\) to be a natural number, \(k\) must be such that \(b(b + 2 - k) = 2k\) has integer solutions. This is true for \(k = b\).

13. Therefore, \(a = 1\) and \(b\) can be any natural number.

The final answer is \( \boxed{ (a, b) = (1, b) } \) for any natural number \(b\).