Solution: Suppose, for contradiction, that $f\left( \frac{1}{n} \right) = \frac{1}{n^2 + 1}$ for all positive integers $n$. Define a function \[ g(z) \coloneqq f(z) - \frac{1}{z^2 + 1}, \quad |z| < 1. \] Since $f$ is analytic in $|z| \leq 1$, and the function $\frac{1}{z^2 + 1}$ is analytic in the open unit disk $|z| < 1$ (its poles are at $z = \pm i$, which lie on $|z| = 1$), it follows that $g$ is analytic in $|z| < 1$. Evaluating $g$ at the points $z = \frac{1}{n}$, we have \begin{align*} g\left( \frac{1}{n} \right) & = f\left( \frac{1}{n} \right) - \frac{1}{\left( \frac{1}{n} \right)^2 + 1} \\ & = \frac{1}{n^2 + 1} - \frac{1}{\frac{1}{n^2} + 1} \\ & = \frac{1}{n^2 + 1} - \frac{1}{\frac{1 + n^2}{n^2}} \\ & = \frac{1}{n^2 + 1} - \frac{n^2}{n^2 + 1} = 0. \end{align*} Thus, $g\left( \frac{1}{n} \right) = 0$ for all $n \in \mathbb{N}$. Since the sequence $\left( \frac{1}{n} \right)$ converges to $0$, which lies in the open unit disk $|z| < 1$, and $0$ is a limit point of the sequence, the identity theorem for analytic functions implies that $g(z) \equiv 0$ in $|z| < 1$. Hence, \[ f(z) = \frac{1}{z^2 + 1}, \quad |z| < 1. \]

However, $f$ is given to be analytic in the closed unit disk $|z| \leq 1$, which includes the boundary $|z| = 1$. The function $\frac{1}{z^2 + 1}$ has poles at $z = \pm i$, where $|z| = |i| = 1$, and these points lie on the boundary of the closed unit disk. So, the function can not be analytic on $\overline{D(0,1)} $ which contradicts the assumption that $f$ is analytic in $|z| \leq 1$. Therefore, the assumption that $f\left( \frac{1}{n} \right) = \frac{1}{n^2 + 1}$ for all $n$ is impossible. Hence, there must exist some positive integer $n$ such that $f\left( \frac{1}{n} \right) \neq \frac{1}{n^2 + 1}$.