6.50 The sum of 50 positive numbers is 231, and the sum of their squares is 2009. Then, the maximum value of the largest number among these 50 numbers is $\qquad$ .
6. 35 .

Let the maximum number be $x$, and the other 49 numbers be $a_{1}, a_{2}$, $\cdots, a_{49}$, and $x \geqslant a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{49}$. Then
$$
\begin{array}{l}
\sum_{i=1}^{49} a_{i}=231-x, \\
2009=x^{2}+\sum_{i=1}^{49} a_{i}^{2} \geqslant x^{2}+\frac{1}{49}\left(\sum_{i=1}^{49} a_{i}\right)^{2} \\
=x^{2}+\frac{1}{49}(231-x)^{2},
\end{array}
$$

i.e., $50 x^{2}-462 x+231^{2} \leqslant 49 \times 2009$.
Solving this, we get $x \leqslant 35$.
Equality holds if and only if $a_{1}=a_{2}=\cdots=a_{49}=4$, in which case $x=35$.