6. Let $x_{1}, x_{2}, \cdots, x_{62}$ all be natural numbers, $x_{1}<x_{2}<\cdots<$ $x_{62}$, and $x_{1}+x_{2}+\cdots+x_{62}=1994$. Then the maximum value that $x_{62}$ can take is ( ).
(A) 101
(B) 102
(C) 103
(D) 104
6. (C).
$$
\begin{array}{l}
x_{1}+\left(x_{2}-1\right)+\left(x_{3}-2\right)+\cdots+\left(x_{62}-61\right) \\
=103,
\end{array}
$$

To maximize $x_{62}$, it is necessary to minimize $x_{1}, x_{2}-1, x_{3}-2, \cdots, x_{61}-60$. For this, we can take $x_{1}=x_{3}-1=x_{3}-2=\cdots=x_{61}-60=1$, then $x_{62}-61=42$. Therefore, the maximum value of $x_{62}$ is 103.