If $n$ is a positive integer such that $8n+1$ is a perfect square, then
$\textbf{(A)}~n\text{ must be odd}$
$\textbf{(B)}~n\text{ cannot be a perfect square}$
$\textbf{(C)}~n\text{ cannot be a perfect square}$
$\textbf{(D)}~\text{None of the above}$
1. Let \( 8n + 1 = k^2 \) for some integer \( k \). This implies:
   \[
   8n = k^2 - 1
   \]
   \[
   8n = (k-1)(k+1)
   \]

2. Since \( 8n \) is even, \( (k-1)(k+1) \) must also be even. Note that \( k-1 \) and \( k+1 \) are two consecutive even numbers, so one of them must be divisible by 2, and the other must be divisible by 4. Therefore, \( (k-1)(k+1) \) is divisible by 8.

3. Let us check if \( n \) must be odd. Suppose \( n \) is even, say \( n = 2m \). Then:
   \[
   8(2m) + 1 = k^2
   \]
   \[
   16m + 1 = k^2
   \]
   This implies \( k^2 \equiv 1 \pmod{16} \). The quadratic residues modulo 16 are \( 0, 1, 4, 9 \), so \( k^2 \equiv 1 \pmod{16} \) is possible. Therefore, \( n \) can be even.

4. Now, let us check if \( n \) can be a perfect square. Suppose \( n = m^2 \). Then:
   \[
   8m^2 + 1 = k^2
   \]
   \[
   k^2 - 8m^2 = 1
   \]
   This is a Pell's equation, which has infinitely many solutions in integers \( (k, m) \). Therefore, \( n \) can be a perfect square.

5. Since \( n \) can be even and \( n \) can be a perfect square, options (A), (B), and (C) are incorrect. Therefore, the correct answer is:
   \[
   \boxed{\text{None of the above}}
   \]