1. Let $n$ ($n<100$) be a positive integer, and there exists a positive integer $k$, such that $1 \leqslant k \leqslant n-1$, satisfying
$$
\frac{4 k+1}{2 n}+\frac{1-2 k^{2}}{n^{2}}=\frac{1}{2} \text {. }
$$

How many values of $n$ satisfy the condition? Prove your conclusion.
1. From equation (1), we get $(2 k-n)^{2}=n+2$.

Solving, we get $k=\frac{1}{2}(n \pm \sqrt{n+2})$.
Since $k$ is an integer, therefore, $n=m^{2}-2(m \in \mathbf{N})$.
Thus, $k=\frac{1}{2}\left(m^{2} \pm m-2\right)$.
Also, $1 \leqslant k \leqslant n-1$, so
$$
1 \leqslant \frac{m^{2} \pm m-2}{2} \leqslant m^{2}-3 \text {. }
$$

Solving, we get $m \geqslant 3$.
But $n<100$, hence the $n$ that satisfies the problem are $7,14,23,34,47,62,79,98$.
A total of 8.