We first use the lemma $\tau(n)<2\sqrt{n}$ to get $n=\tau(n)+2023<2\sqrt{n}+2023$. This gives $n<46^2=2116$. The problem condition implies $n > 2023$, hence $n \geq 2+2023=2025$. In addition, $n= 2025$ fails so $2026 \leq n \leq 2115$. But this means $n$ is not a perfect square, so $\tau(n)$ is even, implying that $n$ is odd. Considering all odd $n$'s varying from $2026$ to $2115$, we can use the $\tau(n)$ formula to get that $\tau(n) \leq 16$. This gives $n \geq 2023+16 = 2039$. Now, directly checking all odd $n$'s from $2027$ to $2039$ gives no such $n$. Hence, there are no positive integers $n$ such that $\tau(n)+2023=n$.

We claim all primes $k>2040$ work. First, if $k \mid n$, write $n = k^a \cdot l$ for some positive integers $a, l$ with $(l, k)=1$. Then the equation is equivalent to $(a+1) \tau(l) + 2023 = k^a \cdot l$. We use the lemma $\tau(l)<2\sqrt{l}$ to get $k^a \cdot l < 2(a+1) \sqrt{l} + 2023$. Now, consider the function $f(x)=k^ax^2-2(a+1)x-2023$ for $x \geq 1$. We get that $f'(x)=2(k^ax-(a+1)) \geq 2(k^a-(a+1))>0,$ so the function is increasing on $[1, \infty)$. This means $f(\sqrt{l}) \geq f(1)=k^a-(2a+2025)>0,$ absurd. This means $(n, k)=1$. The equation is now $2\tau(n)+2023=n$. The condition implies $n > 2023$, hence $n \geq 2+2023=2025$. We now use the lemma $\tau(n)<2\sqrt{n}$ to get $n<4\sqrt{n}+2023$, so $n \leq 2218$. In addition, $n= 2025$ fails and $n$ is odd, so $2027 \leq n \leq 2218$. Considering all odd $n$'s varying from $2026$ to $2218$, we can use the $\tau(n)$ formula to get that $\tau(n) \leq 16$. This gives $n \geq 2023+16 = 2039$. Now, directly checking all odd $n$'s from $2027$ to $2039$ gives $n=2027$ and $n=2031$. Our claim is proved, the problem is solved.