Question: In the most recent national mathematics textbook for 10th grade in my home country, 
 ${a}^{\frac 1n}$ is defined in the following way:

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> For every natural number $n \ge2$, positive number $a$ raised to power of ${\frac 1n}$ is defined as: $${a}^{\frac 1n} = \sqrt[n]a$$
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($a \in \Bbb{R}$ as stated before definition)

and then adds:

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> Throught this book if $a\lt0$, ${a}^{\frac 1n}$ is not defined. For example expressions such as ${(-1)}^{\frac 13}$ and ${(-2)}^{\frac 24}$ are undefined.
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(Definition and comment above can be seen on page 65 of the pdf version available through the following link (Persian)(PDF): <http://chap.sch.ir/sites/default/files/lbooks/96-97/40/C110211.pdf>)

What is the reason of defining it this way?

Answer: 
Presumably, the students do not learn about the complex numbers. As a result, there is no meaningful definition for things like $(-1)^{\frac12}$, which under any reasonable definition should be something that, when squared, gives $-1$; but such a number does not exist within the real numbers.

In case $n$ is odd, there is of course a good definition as a negative number, but for simplicity they presumably just decide to restrict radicals to positive numbers, no matter which radical.
That is for two reasons:

First, $a^x$, for $x\in \mathbf R$ is defined as $\;\mathrm e^{x\ln a}$, and $$\ln a$ is defined for $a>0$.

Second, while we may very well speak of $\;\sqrt[3]{\mathstrut -8}$, for instance, the exponential notation $(-8)^{\tfrac13}$ leads to inconsistencies:
\begin{align}(-8)^{\tfrac13}&=(-8)^{\tfrac26}=\bigl((-8)^2\bigr)^{\tfrac16}=(64)^{\tfrac16}=2,\\
\text{but}\qquad
(-8)^{\tfrac13}&=\Bigl((-8)^{\tfrac16}\Bigr)^2\qquad\text{is not defined!}
\end{align}