10. Betya and Vasya are playing a game. Betya selects 100 (possibly the same) non-negative numbers $x_{1}, x_{2}, \cdots, x_{100}$, whose sum equals 1. Vasya then divides them into 50 pairs, calculates the product of the two numbers in each pair, and writes the largest of these products on the blackboard. Betya wants the number written on the blackboard to be as large as possible, while Vasya wants it to be as small as possible. What will the number written on the blackboard be under the correct strategy?
10. $\frac{1}{396}$.

If Betya selects the numbers $\frac{1}{2}, \frac{1}{198}, \cdots, \frac{1}{198}$, then no matter how Vasya pairs these 100 numbers, the number paired with $\frac{1}{2}$ will always be $\frac{1}{198}$, and their product is $\frac{1}{396}$.

Below is the proof: No matter how Betya selects the numbers, the number Vasya gets will not be greater than $\frac{1}{396}$.
Assume the 100 numbers selected by Betya are ordered as follows:
$$
x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{100} \text {. }
$$

We can pair $x_{k}$ with $x_{101-k}$. Thus, the products of the pairs are
$$
x_{1} x_{100}, x_{2} x_{99}, \cdots, x_{k} x_{101-k}, \cdots, x_{50} x_{51} \text {. }
$$

For $1 \leqslant k \leqslant 49$, let $a=x_{k} x_{101-k}$.
By $x_{k} \leqslant x_{k-1} \leqslant \cdots \leqslant x_{1}$, we have
$$
k x_{k} \leqslant x_{1}+x_{2}+\cdots+x_{k} \text {. }
$$

Thus, $k a=k x_{k} x_{101-k} \leqslant\left(x_{1}+x_{2}+\cdots+x_{k}\right) x_{101-k}$.
By $x_{101-k} \leqslant x_{100-k} \leqslant x_{99-k} \leqslant \cdots \leqslant x_{k+1}$, we have
$$
\begin{array}{l}
(101-2 k) x_{101-k} \\
\leqslant x_{101-k}+x_{100-k}+\cdots+x_{k+1} \\
\leqslant x_{k+1}+x_{k+2}+\cdots+x_{100} \\
=1-x_{1}-x_{2}-\cdots-x_{k} .
\end{array}
$$

From the above two inequalities, we get
$$
\begin{array}{l}
k(101-2 k) a \\
\leqslant\left(x_{1}+x_{2}+\cdots+x_{k}\right)\left(1-x_{1}-x_{2}-\cdots-x_{k}\right) \\
=x(1-x),
\end{array}
$$

where $x=x_{1}+x_{2}+\cdots+x_{k}$.
By the AM-GM inequality, we have
$$
x(1-x) \leqslant\left(\frac{x+(1-x)}{2}\right)^{2}=\frac{1}{4} \text {. }
$$

Noting that, for $1 \leqslant k \leqslant 49$,
$$
\begin{array}{l}
(k-1)(99-2 k) \geqslant 0 \\
\Rightarrow k(101-2 k) \geqslant 99 .
\end{array}
$$

Thus, $x_{k} x_{101-k}=a$
$$
\leqslant \frac{1}{4 k(101-2 k)} \leqslant \frac{1}{396} \text {. }
$$

Since $x_{51} \leqslant x_{50} \leqslant \cdots \leqslant x_{1}$, we have
$$
\begin{array}{l}
x_{50} \leqslant \frac{x_{1}+x_{2}+\cdots+x_{50}}{50} \leqslant \frac{1}{50}, \\
x_{51} \leqslant \frac{x_{1}+x_{2}+\cdots+x_{50}+x_{51}}{51} \leqslant \frac{1}{51} .
\end{array}
$$

Therefore, $x_{50} x_{51} \leqslant \frac{1}{50 \times 51}<\frac{1}{396}$.