Let us consider the given problem and analyze the conditions set by the equations:

1. $\sum_{k=1}^n a_k = 0$,
2. $\sum_{k=1}^n a_k^2 = 1$,
3. $\sqrt{n} \cdot \left(\sum_{k=1}^n a_k^3\right) = 2(b\sqrt{n} - 1)$, where $b = \max_{1 \leq k \leq n} \{a_k\}$.

The task is to find all integers $n \geq 2$ for which these real numbers $a_k$ exist.

### Step 1: Analyze Sum Constraints

Given $\sum_{k=1}^n a_k = 0$, the sum of the numbers $a_k$ is zero. This indicates a balance between positive and negative terms among $a_k$.

### Step 2: Analyze Maximum Constraint

Let $b = \max_{1 \leq k \leq n} \{a_k\}$. From the constraint $\sqrt{n} \cdot \left(\sum_{k=1}^n a_k^3\right) = 2(b\sqrt{n} - 1),$ we must have:
$$\sum_{k=1}^n a_k^3 = \frac{2b\sqrt{n} - 2}{\sqrt{n}} = 2b - \frac{2}{\sqrt{n}}.$$

### Step 3: Consider Equality Conditions

We now explore whether the equality conditions can hold for different values of $n$:

- For $n = 2$: Consider $a_1 = b$ and $a_2 = -b$. Then:
$$\sum_{k=1}^2 a_k = b - b = 0,$$
$$\sum_{k=1}^2 a_k^2 = b^2 + b^2 = 2b^2 = 1 \Rightarrow b^2 = \frac{1}{2},$$
$$\sum_{k=1}^2 a_k^3 = b^3 - b^3 = 0,$$
$$\sqrt{2} \cdot (0) = 2(b \sqrt{2} - 1) \Rightarrow b \sqrt{2} = 1.$$

This gives $b = \frac{1}{\sqrt{2}}$, which satisfies all conditions for $n = 2$.

### Step 4: Generalize for Even $n > 2$

Considering $n$ being even:
- Let $a_k = \frac{1}{\sqrt{n}}$ for $k$ in the first half and $a_k = -\frac{1}{\sqrt{n}}$ for $k$ in the second half.
- Then:
$$\sum_{k=1}^n a_k = 0,$$
$$\sum_{k=1}^n a_k^2 = \frac{n}{n} = 1,$$
$$\sum_{k=1}^n a_k^3 = \frac{n/2}{(\sqrt{n})^3} - \frac{n/2}{(\sqrt{n})^3} = 0,$$
$$\sqrt{n} \cdot 0 = 2\left(\frac{1}{\sqrt{n}} \sqrt{n} - 1\right) = 0.$$

Hence, the conditions are satisfied for all even $n$.

### Conclusion:

The integers $n$ for which the conditions are met are all even integers $n \geq 2$. Thus, the final answer is:
$$\boxed{\text{even integers } n \geq 2}$$