Define $b_k=a_k+a_{k+1}$. We have $b_0=-\frac 1n$. Also, $$b_{k-1}\cdot b_k=b_{k-1}-b_k\Rightarrow b_k(b_{k-1}+1)=b_{k-1}\Rightarrow b_k=\frac{b_{k-1}}{b_{k-1}+1}$$Hence, $b_1=-\frac 1{n-1}, b_2=-\frac 1{n-2},\cdots ,b_{n-1}=-1$. Then, if $N\ge n+1$, then $b_n$ exist. But, $b_n(b_{n-1}+1)=b_{n-1}$ and $b_{n-1}=-1$. This gives us $0=-1$, contradiction. Therefore, $N\leq n$. Also, $b_i=-\frac 1{n-i}$ works for $N=n$. (For this, you can set $a_0=0$ and $a_{i+1}=-\frac 1{n-i}-a_i$ for $i\in [0,n-1]$) So the answer is $max \{N\}=n$.

Condition (2) rewrites as $a_{k+1} = 1 - a_k - \frac{1}{a_{k-1} + a_k + 1}$. In particular, the entire sequence is uniquely defined by $a_0$. Claim: $a_k + a_{k+1} = -\frac{1}{n-k}$ for all $0\le k\le n-1$. Proof: We proceed with induction on $k$. The base case of $k = 0$ is condition (1). Induction step: suppose for some $0\le m\le n-2$ that the claim is true for $k = m$. Then we have $$a_{m+1} + a_{m+2} = a_{m+1} + 1 - a_{m+1} - \frac{1}{1 - \frac{1}{m-k}} = -\frac{1}{n - (m+1)}$$ as desired, and the induction step is complete. The claim implies that $N\le n$, and that equality is achievable for any value of $a_0$.

We claim the answer is $\boxed{n}$. The key claim is that $a_k+a_{k+1}=\frac{1}{k-n}$. Proof: We induct on $k$. Base cases: Clearly $k=0$. Now we have $-\frac{a_1+a_2}{n}=a_0-a_2=-\frac{1}{n}-a_1-a_2$. So $a_1+a_2=n\left(\frac{1}{n}+a_1+a_2\right)=1+n(a_1+a_2)$. So $a_1+a_2=\frac{1}{1-n}$. Thus, $k=1$ is also a valid base case. Inductive step: Suppose $a_k+a_{k-1}=\frac{1}{k-1-n}$. Then we have \[\frac{a_k+a_{k+1}}{k-1-n}=a_{k-1}-a_{k+1}=\frac{1}{k-1-n}-a_k-a_{k+1}\] So we multiply both sides by $k-1-n$ and get $a_k+a_{k+1}=1-(k-1-n)(a_k+a_{k+1})$, so \[(k-n)(a_k+a_{k+1})=1\implies a_k+a_{k+1}=\frac{1}{k-n},\]so the induction is complete. $\blacksquare$ If $N>n$, then $a_n+a_{n+1}=\text{undefined}$, so $N\le n$. For equality, set $a_0=0$ and $a_{i+1}=\frac{1}{i-n}-a_i$ for $0\le i\le n-1$.

Nothing too different from the rest, but: We notice, that $(a_k+a_{k-1})(a_k+a_{k+1})=a_{k-1}-a_{k+1} \Leftrightarrow (a_k+a_{k-1})(a_k+a_{k+1})= (a_k+a_{k-1}) - (a_k+a_{k+1}) \Leftrightarrow \frac{1}{(a_k+a_{k+1})} - \frac{1}{(a_k+a_{k-1})} = 1$. Combined with how suggestive $a_0+a_1 = -\frac{1}{n}$ looks, we are motivated to let $b_k = \frac{1}{(a_k+a_{k+1})}$. Then $b_0 = -n$ and $b_k = b_{k-1}+1$ and hence $b_k = k-n$. Clearly for all $N>n$, we get $b_n = n -n =0$. But this is absurd as $b_i$'s are all reciprocals of real numbers. So $n \le N$.

The second condition rewrites to $a_{k+1}(1+a_{k-1}+a_k)=a_{k-1}-a_{k-1}a_k-a_k^2$. Claim: $a_k+a_{k+1}=-\tfrac{1}{n-k}$ for $0\leq k\leq n-1$. Proof. We will use induction on $k$. The base case of $k=0$ is implied by the problem statement. Suppose we have proven the inductive hypothesis for some $k-1$. We will show it is true for $k$ as well. Indeed, we have \begin{align*} a_{k+1}&=\frac{a_{k-1}-a_{k-1}a_k-a_k^2}{a_{k-1}+a_k+1}=\frac{\left(-\frac{1}{n-k+1}-a_k\right)-\left(-\frac{1}{n-k+1}-a_k\right)(a_k)-a_k^2}{1-\frac{1}{n-k+1}} \\ &=\frac{-a_k\left(1-\frac{1}{n-k+1}\right)-\frac{1}{n-k+1}}{1-\frac{1}{n-k+1}} \\ &=-a_k-\frac{1}{n-k+1}\cdot\frac{n-k+1}{n-k}=-a_k-\frac{1}{n-k}. \end{align*}Therefore, $a_k+a_{k+1}=-\tfrac{1}{n-k}$, as desired, which completes the inductive step. When $a_{n-1}+a_n=-1$, we have \begin{align*} a_{n-1}-a_{n-1}a_n-a_n^2=(-1-a_n)-(-1-a_n)a_n-a_n^2=-1, \end{align*}so we have $a_{n+1}\cdot 0=-1$, meaning that $a_{n+1}$ is not well-defined. Hence, the maximum value of $N$ is $n$, and it is easy to give a construction for a valid sequence of length $n$. Remark: I first tried looking at the values of $a_k+a_{k+1}$ that would make the sequence "stop", and I found that they were $-1$ and $-\tfrac{1}{2}$, but this didn't help me narrow down when this would happen. Then, I started computing some terms of the sequence and I found \begin{align*} a_1=a_0-\frac{1}{n}, a_2=a_0-\frac{1}{n-1}+\frac{1}{n}, a_3=-a_0-\frac{1}{n}+\frac{1}{n-1}-\frac{1}{n-2}, \end{align*}which led me to introduce my claim, and finish.

Same recurrence relation as: https://artofproblemsolving.com/community/c6h401066p2233182 Serbia MO 2011

Very similar to other solutions, but for storage: Let $b_k=a_k+a_{k-1}$, then we get that $b_kb_{k+1}=b_k-b_{k+1}$ by the second condition, so $b_{k+1}=\frac{b_k}{b_k+1}$. By induction it is clear that $b_{k}=-\frac{1}{n+1-k}$, so $b_{n}=-1$. Then $b_{n+1}$ is undefined, so the answer is $\boxed{n}$

I just did the simplest thing one would do and I was done, are you kidding me EGMO? If $a_0+a_1=-\frac{1}{n}$ putting this into the formula $(2)$ we get $a_1+a_2=-\frac{1}{n-1}$.So we choose this decreasing property and induct to get $a_{k}+a_{k+1}=-\frac{1}{n-k}$.So if we set $b_i=a_i+a_{i+1}$ we get that $b_i$ for $i>n-1$ is undefined.So the maximum possible $N$ is $n$ itself and any construction for $a_0$ works.

This problem was proposed by Călin Popescu, from Romania.