The answer is 6. We will show that the desired integers $n$ are precisely the odd positive integers less than or equal to 11.

We observe that $2022+\frac{1}{2}=\frac{4045}{2}$ and $25+\frac{1}{2}=\frac{51}{2}$, so we are looking for the positive integers $n$ for which the fraction $\frac{4045^{n}+51^{n}}{2^{n}}$ is an integer.

We first observe that if $n=2k$ is even this never happens: indeed the numerator is even, but not divisible by 4, since $4045^{2k}$ and $51^{2k}$ both leave remainder 1 upon division by 4 (to prove this fact it suffices to observe that $4045^{2k}-1=\left(4045^{k}+1\right)\left(4045^{k}-1\right)$ is a product of two even numbers, and is therefore divisible by 4; similarly for $51^{2k}-1$), and so their sum leaves remainder 2 upon division by 4.

When $n=2k+1$ is odd, instead, we can use the well-known factorization for a sum of two odd powers with the same exponent to obtain
$$4045^{2k+1}+51^{2k+1}=(4045+51) \cdot\left(4045^{2k}-4045^{2k-1} \cdot 51+\cdots-4045 \cdot 51^{2k-1}+51^{2k}\right)$$
The number in the second parenthesis is the algebraic sum of $2k+1$ odd terms, and is therefore odd. The product just written is therefore divisible by $2^{2k+1}$ if and only if $4045+51=4096=2^{12}$ is. This happens if and only if $2k+1 \leq 12$, that is, for $k=0,1, \ldots, 5$, values which give the 6 desired solutions.