4. Find all natural numbers $n \geqslant 2$ for which there exist mutually distinct real numbers $a_{1}, a_{2}, \ldots, a_{n}$ such that the following two conditions are satisfied:

- when each of these $n$ numbers is replaced by the sum of the remaining $n-1$ numbers (all except it), the same set of numbers is obtained;
- there is no partition of these $n$ numbers into two non-empty disjoint sets such that the sums of the $(n+1)$-th powers of the elements of these sets are equal.
4. For $n=2$, any two-element set satisfies the conditions of the problem. Let us further assume $n>2$.

Let $S=a_{1}+a_{2}+\cdots+a_{n}$. From the first condition, we have $\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}=$ $\left\{S-a_{1}, S-a_{2}, \ldots, S-a_{n}\right\}$, so the sums of the elements from these two sets are equal. We get $S=n S-S$, i.e., $S(n-2)=0$, from which it follows that $S=0$. Therefore, $\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}=\left\{-a_{1},-a_{2}, \ldots,-a_{n}\right\}$. Note, if for some $i, 1 \leqslant i \leqslant n$, we have $a_{i}=-a_{i}$, it would follow that $a_{i}=0$, so we can have at most one such $i$. Without loss of generality, let $a_{1}=-a_{2}$. Then we also have $a_{2}=-a_{1}$, so the initial set of numbers (excluding zero, in the case where 0 belongs to the initial set) can be divided into pairs $\{x,-x\}, x \neq 0$.

Consider first the case $n=2 k$, for $k \geqslant 2$. Then we have $k$ pairs of the form $\left\{x_{i},-x_{i}\right\}, x_{i} \neq 0$, for $i=1,2, \ldots, k$. We can divide them into two disjoint sets $\left\{x_{1},-x_{1}\right\}$ and $\left\{x_{2},-x_{2}, \ldots, x_{k},-x_{k}\right\}$, where $x_{1}^{n+1}+\left(-x_{1}\right)^{n+1}=0=x_{2}^{n+1}+$ $\left(-x_{2}\right)^{n+1}+\cdots+x_{k}^{n+1}+\left(-x_{k}\right)^{n+1}$. Therefore, such $n$ are not solutions to the problem due to the second condition.

Now consider the case $n=2 k+1$, for $k \geqslant 1$. We conclude that one of the observed numbers is equal to 0, and the others form pairs as noted above. The given set can be divided into two disjoint sets $\left\{0, x_{1}, x_{2}, \ldots, x_{k}\right\}$ and $\left\{-x_{1},-x_{2}, \ldots,-x_{k}\right\}$, where $0^{n+1}+x_{1}^{n+1}+x_{2}^{n+1}+\cdots+x_{k}^{n+1}=\left(-x_{1}\right)^{n+1}+$ $\left(-x_{2}\right)^{n+1}+\cdots+\left(-x_{n}\right)^{n+1}$. Therefore, such $n$ are also not solutions to the problem due to the second condition.

Thus, the only solution to the problem is $n=2$.