Three. (25 points) If the equation with respect to $x$
$$
x^{2}-34 x+34 k-1=0
$$

has at least one positive integer root, find the value of the positive integer $k$ that satisfies the condition.
Three, let the two roots of the equation be $x_{1}$ and $x_{2}$, and $x_{1}$ is a positive integer.
Then $x_{1}+x_{2}=34, x_{1} x_{2}=34 k-1$.
Thus, $x_{2}=34-x_{1}$ is also an integer.
Since $k$ is a positive integer and $x_{1} x_{2}=34 k-1$, we know $x_{2}>0$.
Therefore, $x_{2}$ is a positive integer.
Notice,
$$
\begin{array}{l}
\left(x_{1}+1\right)\left(x_{2}+1\right)=x_{1} x_{2}+x_{1}+x_{2}+1=34(k+1) \\
\Rightarrow 17 \mid\left(x_{1}+1\right)\left(x_{2}+1\right) \\
\Rightarrow 17 \mid\left(x_{1}+1\right) \text { or } 17 \mid\left(x_{2}+1\right) .
\end{array}
$$

If $17 \mid\left(x_{1}+1\right)$, then
$$
\begin{array}{l}
x_{1}+1 \leqslant x_{1}+x_{2}=34 \\
\Rightarrow x_{1}+1=17 \text { or } x_{1}+1=34 . \\
\text { When } x_{1}+1=17 \text {, } x_{1}=16, x_{2}=18, \text { at this time, } \\
34 k-1=16 \times 18,
\end{array}
$$
$k$ has no integer solution;
When $x_{1}+1=34$, $x_{1}=33, x_{2}=1$, at this time,
$$
34 k-1=33 \times 1 \Rightarrow k=1 \text {. }
$$

If $17 \mid \left(x_{2}+1\right)$, similarly we get $k=1$.
In summary, the positive integer $k=1$ satisfies the condition.