10. Non-negative real numbers $a_{i}(i=1,2, \cdots, n)$, satisfy: $a_{1}+a_{2}+a_{3}+\cdots+a_{n}=1$, find the minimum value of $\frac{a_{1}}{1+a_{2}+\cdots+a_{n}}+\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}}+\cdots+\frac{a_{n}}{1+a_{1}+a_{2}+\cdots+a_{n-1}}$.
10. Solution: By adding a constant 1 to each fraction in the required expression, and after finding a common denominator, replace $a_{1}+a_{2}+\cdots+a_{n}$ with the constant 1, we get
$$
\frac{a_{1}}{1+a_{2}+\cdots+a_{n}}+1=\frac{1+\left(a_{1}+a_{2}+\cdots+a_{n}\right)}{2-a_{1}}=\frac{2}{2-a_{1}}
$$

Similarly, we have $\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}}+1=\frac{2}{2-a_{2}}, \cdots, \frac{a_{n}}{1+a_{1}+a_{2}+\cdots+a_{n-1}}+1=\frac{2}{2-a_{n}}$.
Let $y=\frac{a_{1}}{1+a_{2}+\cdots+a_{n}}+\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}}+\cdots+\frac{a_{n}}{1+a_{1}+a_{2}+\cdots+a_{n-1}}$.
Then $y+n=\frac{2}{2-a_{1}}+\frac{2}{2-a_{2}}+\cdots+\frac{2}{2-a_{n}}$.
To use the Cauchy inequality, note that
$$
\left(2-a_{1}\right)+\left(2-a_{2}\right)+\cdots+\left(2-a_{n}\right)=2 n-\left(a_{1}+a_{2}+\cdots+a_{n}\right)=2 n-1 .
$$

Thus, $(2 n-1)\left(\frac{1}{2-a_{1}}+\frac{1}{2-a_{2}}+\cdots+\frac{1}{2-a_{n}}\right)$
$$
\begin{array}{l}
=\left[\left(2-a_{1}\right)+\left(2-a_{2}\right)+\cdots+\left(2-a_{n}\right)\right] \cdot\left(\frac{1}{2-a_{1}}+\frac{1}{2-a_{2}}+\cdots+\frac{1}{2-a_{n}}\right) \\
\geqslant\left[\sqrt{2-a_{1}} \cdot \frac{1}{\sqrt{2-a_{1}}}+\sqrt{2-a_{2}} \cdot \frac{1}{\sqrt{2-a_{2}}}+\cdots+\sqrt{2-a_{n}} \cdot \frac{1}{\sqrt{2-a_{n}}}\right]^{2}=n^{2} .
\end{array}
$$

Therefore, $y+n \geqslant \frac{2 n^{2}}{2 n-1}, y \geqslant \frac{2 n^{2}}{2 n-1}-n=\frac{n}{2 n-1}$.
Equality holds if and only if $a_{1}=a_{2}=\cdots=a_{n}=\frac{1}{n}$, thus the minimum value of $y$ is $\frac{n}{2 n-1}$.