10. (32nd Russian Mathematical Olympiad) For which positive integers $n$, can two non-integer positive rational numbers $a, b$ be found such that $a+b$ and $a^{n}+b^{n}$ are both integers?
10. All odd numbers.

If the positive integer $n$ is odd, then it suffices to let
$$
a=\frac{1}{2}, b=\frac{2^{n}-1}{2} \text {. }
$$

In fact, $a+b$ is an integer, and
$$
a^{n}+b^{n}=(a+b)\left(a^{n-1}-a^{n-2} b+\cdots+b^{n-1}\right)=2^{n-1}\left(a^{n-1}-a^{n-2} b+\cdots+b^{n-1}\right) \text {. }
$$

Since the denominator of each term in the parentheses is $2^{n-1}$, $a^{n}+b^{n}$ is also an integer.
Now suppose the positive integer $n$ is even, i.e., $n=2 k\left(k \in \mathbf{N}_{+}\right)$.
If there exist the required positive rational numbers $a, b$, then since $a+b$ is an integer, their irreducible fraction representations have the same denominator, i.e., $a=\frac{p}{d}, b=\frac{q}{d}$, and $p+q$ is divisible by $d$. At the same time, we have
$$
p^{n}+q^{n}=\left(p^{2 k}-q^{2 k}\right)+2 q^{2 k}=\left(p^{2}-q^{2}\right)\left(p^{2 k-2}+p^{2 k-4} q^{2}+\cdots+q^{2 k-2}\right)+2 q^{2 k}=(p+q) K+2 q^{2 k} \text {, }
$$

where $K=(p-q)\left(p^{2 k-2}+p^{2 k-4} q^{2}+\cdots+q^{2 k-2}\right)$ is an integer.
Note that $a^{n}+b^{n}=\frac{p^{n}+q^{n}}{d^{m}}$ is also an integer, i.e., $p^{n}+q^{n}$ is divisible by $d^{n}$, and in particular, by $d$.
Since $d \mid(p+q)$, it follows that $d \mid 2 q^{\prime \prime}$.
But $\frac{q}{d}$ is an irreducible fraction, so $d \mid 2$, i.e., $d=2$.
Thus, $p^{n}, q^{n}$ are both squares of odd numbers, and their remainders when divided by 4 are both 1.
Therefore, $p^{n}+q^{n}$ cannot be divisible by 4.
However, as previously shown, $p^{n}+q^{n}$ is divisible by $d^{n}=2^{n}=2^{2 k}$, which is clearly a multiple of 4. This is a contradiction.