Find all integers $n \geq 2$ for which there exist the real numbers $a_{k}, 1 \leq k \leq n$, which are satisfying the following conditions:

$$
\sum_{k=1}^{n} a_{k}=0, \sum_{k=1}^{n} a_{k}^{2}=1 \text { and } \sqrt{n} \cdot\left(\sum_{k=1}^{n} a_{k}^{3}\right)=2(b \sqrt{n}-1), \text { where } b=\max _{1 \leq k \leq n}\left\{a_{k}\right\}
$$
We have: $\left(a_{k}+\frac{1}{\sqrt{n}}\right)^{2}\left(a_{k}-b\right) \leq 0 \Rightarrow\left(a_{k}^{2}+\frac{2}{\sqrt{n}} \cdot a_{k}+\frac{1}{n}\right)\left(a_{k}-b\right) \leq 0 \Rightarrow$

$$
a_{k}^{3} \leq\left(b-\frac{2}{\sqrt{n}}\right) \cdot a_{k}^{2}+\left(\frac{2 b}{\sqrt{n}}-\frac{1}{n}\right) \cdot a_{k}+\frac{b}{n} \forall k \in\{1,2, \cdots, n\} \cdot(k)
$$

Adding up the inequalities ( $k$ ) we get

$$
\begin{gathered}
\sum_{k=1}^{n} a_{k}^{3} \leq\left(b-\frac{2}{\sqrt{n}}\right) \cdot\left(\sum_{k=1}^{n} a_{k}^{2}\right)+\left(\frac{2 b}{\sqrt{n}}-\frac{1}{n}\right) \cdot\left(\sum_{k=1}^{n} a_{k}\right)+b \leftrightarrow \\
\sum_{k=1}^{n} a_{k}^{3} \leq b-\frac{2}{\sqrt{n}}+b \leftrightarrow \sqrt{n} \cdot\left(\sum_{k=1}^{n} a_{k}^{3}\right) \leq 2(b \sqrt{n}-1) .
\end{gathered}
$$

But according to hypothesis,

$$
\sqrt{n} \cdot\left(\sum_{k=1}^{n} a_{k}^{3}\right)=2(b \sqrt{n}-1) .
$$

Hence it is necessarily that

$$
\begin{gathered}
a_{k}^{3}=\left(b-\frac{2}{\sqrt{n}}\right) \cdot a_{k}^{2}+\left(\frac{2 b}{\sqrt{n}}-\frac{1}{n}\right) \cdot a_{k}+\frac{b}{n} \forall k \in\{1,2, \cdots, n\} \leftrightarrow \\
\left(a_{k}+\frac{1}{\sqrt{n}}\right)^{2}\left(a_{k}-b\right)=0 \forall k \in\{1,2, \cdots, n\} \leftrightarrow a_{k} \in\left\{-\frac{1}{\sqrt{n}}, b\right\} \forall k \in\{1,2, \cdots, n\}
\end{gathered}
$$

We'll prove that $b>0$. Indeed, if $b \leq 0$.
If $a_{k}=-\frac{1}{\sqrt{n}} \forall k \in\{1,2, \cdots, n\}$ then $\sum_{k=1}^{n} a_{k}=-\sqrt{n} \leq 0$, which is absurd. Hence $\exists m \in\{1,2, \cdots, n-1\}$
such that among the numbers $a_{k}$ we have $n-m$ equal to $-\frac{1}{\sqrt{n}}$ and $m$ equal to $b$. We get $\left\{\begin{array}{l}-\frac{n-m}{\sqrt{n}}+m b=0 \\ \frac{n-m}{n}+m b^{2}=1\end{array}\right.$.
From here, $b=\frac{n-m}{m \sqrt{n}} \Rightarrow \frac{n-m}{n}+\frac{(n-m)^{2}}{m n}=1 \Rightarrow$
$\Rightarrow n-m=m \Rightarrow m=\frac{n}{2}$. Hence $n$ is even.
Conversely, for any even integer $n \geq 2$ we get that there exist the real numbers $a_{k}, 1 \leq k \leq n$, such that

$$
\sum_{k=1}^{n} a_{k}=0, \sum_{k=1}^{n} a_{k}^{2}=1 \text { and } \sqrt{n} \cdot\left(\sum_{k=1}^{n} a_{k}^{3}\right)=2(b \sqrt{n}-1), \text { where } b=\max _{1 \leq k \leq n}\left\{a_{k}\right\}
$$

(We may choose for example $a_{1}=\cdots=a_{\frac{n}{2}}=-\frac{1}{\sqrt{n}}$ and $a_{\frac{n}{2}+1}=\cdots=a_{n}=\frac{1}{\sqrt{n}}$ ).