Answer: $\gamma \ge n$.

At first we will show that for $0 < \gamma < n$ there exists a set $x_0, x_1, \dots, x_n$, for which the inequality from the statement of the problem is not held.

Let $x_0 = x^{-n}$, $x_1 = x_2 = \dots = x_n = x$ for some $x > 0$. Then we have
$$\begin{gathered} x_0^\gamma + x_1^\gamma + \dots + x_n^\gamma = x^{-n\gamma} + n x^\gamma, \\ \frac{1}{x_0} + \frac{1}{x_1} + \dots + \frac{1}{x_n} = x^n + n x^{-1}. \end{gathered}$$
For $\gamma < n$ we want to find $x > 0$ for which $x^n + n x^{-1} > x^{-n\gamma} + n x^\gamma$. For this purpose it is enough to choose $x > 1$: $x^n + n x^{-1} > x^n > (n + 1)x^\gamma > n x^\gamma + x^{-n\gamma}$, i.e. $x^{-n\gamma} > n + 1$ also $x > \sqrt[n]{n+1}$. From here we find a corresponding example.

Let now $\gamma = n$ then by the Cauchy's inequality we receive
$$\begin{gathered} \frac{x_0^n + x_1^n + \dots + x_{n-1}^n}{n} \ge x_0 x_1 \dots x_{n-1} = \frac{1}{x_n}, \\ \frac{x_0^n + x_1^n + \dots + x_{n-2}^n + x_n^n}{n} \ge x_0 x_1 \dots x_{n-2} x_n = \frac{1}{x_{n-1}}, \dots, \end{gathered}$$
$$\frac{x_1^n + x_2^n + \dots + x_n^n}{n} \geq x_1 x_2 \dots x_n = \frac{1}{x_0}.$$
If all these inequalities are added then we will receive the required inequality.

Now we consider the case $\gamma > n$, if we prove that for any set of positive numbers $x_0, x_1, \dots, x_n$ such that $x_0 x_1 \dots x_n = 1$ the inequality $x_0^\gamma + x_1^\gamma + \dots + x_n^\gamma \ge x_0^n + x_1^n + \dots + x_n^n$ holds, then we will receive a necessary inequality for $\gamma$ if we use the already proved inequality for $n$:
$$x_0^\gamma + \dots + x_n^\gamma \ge x_0^n + \dots + x_n^n \Leftrightarrow \\ x_0^\gamma + \dots + x_n^\gamma \ge x_0^{n+\frac{\gamma-n}{n+1}} x_1^{\frac{\gamma-n}{n+1}} \dots x_n^{\frac{\gamma-n}{n+1}} + \dots + x_0^{\frac{\gamma-n}{n+1}} \dots x_{n-1}^{\frac{\gamma-n}{n+1}} x_n^{n+\frac{\gamma-n}{n+1}}.$$
To prove the last inequality we will take advantage of a weighted Cauchy's inequality
$$x_0^{n+\frac{\gamma-n}{n+1}} x_1^{\frac{\gamma-n}{n+1}} \dots x_n^{\frac{\gamma-n}{n+1}} \le \frac{n+\frac{\gamma-n}{n+1}}{\gamma} x_0^{\gamma} + \frac{\gamma-n}{\gamma(n+1)} x_1^{\gamma} + \dots + \frac{\gamma-n}{\gamma(n+1)} x_n^{\gamma}.$$
We will write down similar inequalities for every item, we will add them and we will receive the requisite evidence.