Does $\lim_{N\rightarrow \infty} \sum_{n = 1}^{N} \frac{1}{(N+1) \ln (N+1) - n \ln n } = 1$?

You are right: the limit is $1$. Here it is the full proof:

For all $n \in \{1, \dots N \}$ we have the following inequality: $$(N+1) \log (N+1)- n \log n \ge (N+1) \log (N+1)- n \log (N+1) =\\ = \log(N+1) (N+1-n)$$

Thus $$\sum_{n=1}^N \frac{1}{(N+1)\log(N+1)-n \log n} \le \sum_{n=1}^N \frac{1}{\log(N+1) (N+1-n)} = \frac{H_{N}}{\log(N+1)}$$

where $H_N$ denotes the $N$-th harmonic number.

Since $$\lim_{N \to + \infty} \frac{H_N}{\log(N+1)} =1$$ we have the estimate $$\limsup_{N \to + \infty} \sum_{n=1}^N \frac{1}{(N+1)\log(N+1)-n \log n} \le 1$$ in other words, if the limit exists, it is smaller or equal than $1$.

Proving the other inequality is harder.

Denote by $f(x)=x \log x$. We can compute its derivative $$f'(x)= \log x +1$$

By the Mean Value Theorem, for all $n \in \{1, \dots N \}$ we have $$\frac{(N+1)\log(N+1)-n \log n}{N+1-n}= \frac{f(N+1)-f(n)}{(N+1)-n} = f'(c_n) =\log c_n +1 \le \log(N+1)+1$$ where $c_n$ is some real number in the interval $(n, N+1)$

Thus we have the estimate $$\sum_{n=1}^N \frac{1}{(N+1)\log(N+1)-n \log n} \ge \sum_{n=1}^N \frac{1}{N+1-n} \cdot \frac{1}{\log(N+1)+1} = \frac{H_{N+1}}{\log(N+1)+1}$$ where $H_{N+1}$ denotes the ($N+1$)-th harmonic number.

Since $$\lim_{N \to + \infty} \frac{H_{N+1}}{\log(N+1)+1} =1$$ we have the estimate $$\liminf_{N \to + \infty} \sum_{n=1}^N \frac{1}{(N+1)\log(N+1)-n \log n} \ge 1$$

In other words, we have proved the other inequality: and the limit is indeed $1$.