8. Given $x, y \in \mathbf{R}$, for any $n \in \mathbf{Z}_{+}$, $n x+\frac{1}{n} y \geqslant 1$. Then the minimum value of $41 x+2 y$ is $\qquad$
8.9.

Let the line $l_{n}: n x+\frac{1}{n} y=1$, and call $l_{n} 、 l_{n+1}$ two adjacent lines. Then the intersection point of the two lines is
$$
A_{n}\left(\frac{1}{2 n+1}, \frac{n^{2}+n}{2 n+1}\right) \text {. }
$$

If the intersection point of the line $x+y=1$ and the line $y=0$ is denoted as $A_{0}(1,0)$, then the set of points $\left\{A_{n}\right\}(n \in \mathbf{N})$ connected in sequence forms a polygonal line that encloses the feasible region.
It is easy to prove that the points $A_{n}$ are all above $l_{m}(m>n, m \in \mathbf{N})$.
In this problem, let $41 x+2 y=z$, then $y=-\frac{41}{2} x+\frac{1}{2} z$.
Since $-\frac{41}{2} \in\left[-4^{2},-5^{2}\right]$, when the line $y=-\frac{41}{2} x+\frac{1}{2} z$ passes through the point $A_{4}\left(\frac{1}{9}, \frac{20}{9}\right)$,
$$
z=41 x+2 y=41 \times \frac{1}{9}+2 \times \frac{20}{9}=9 .
$$