We are tasked with finding all integers $n \geq 3$ for which there exist real numbers $a_1, a_2, \ldots, a_{n+2}$ such that $a_{n+1} = a_1$, $a_{n+2} = a_2$, and the recurrence relation

$$a_i a_{i+1} + 1 = a_{i+2}$$

holds for $i = 1, 2, \ldots, n$.

### Step-by-step Solution:

1. Initial Observation:

Notice that we need the sequence to eventually repeat because of the conditions $a_{n+1} = a_1$ and $a_{n+2} = a_2$. This suggests that the cycle of the sequence must be of length dividing $n$.

2. Periodic Nature:

To achieve $a_{n+1} = a_1$ and $a_{n+2} = a_2$, let's assume a cycle length of 3, meaning $a_{k+3} = a_k$ for all $k$. We test $n$ such that $n$ is a multiple of 3 to see if the recurrence can close and repeat.

3. Cycle Test and Closure:

Assume $a_1, a_2, a_3$ exist such that the recurrence relation closes:

$$a_1 a_2 + 1 = a_3,$$
$$a_2 a_3 + 1 = a_4 = a_1,$$
$$a_3 a_4 + 1 = a_5 = a_2.$$

Rewriting the third equation using $a_4 = a_1$, we get:

$$a_3 a_1 + 1 = a_2.$$

We now have a system:

$$a_1 a_2 + 1 = a_3,$$
$$a_2 a_3 + 1 = a_1,$$
$$a_3 a_1 + 1 = a_2.$$

This shows consistency if the cycle length modulo conditions are satisfied.

4. **Conclusion for $n$:**

Since $a_1, a_2, a_3$ corresponds to a cycle that repeats every 3 steps, $n$ must align such that it is a multiple of 3. Thus, the length of the sequence $n$ must be in the form

$$n = 3k,$$

where $k \geq 1$.

This requirement implies the integers $n$ for which such a sequence can exist are precisely multiples of 3.

5. Final Result:

Therefore, the integers $n \geq 3$ that satisfy the existence conditions for the given sequence are:

$$\boxed{3k \text{ for integers } k \geq 1}.$$

This solution confirms the cyclic requirements of the sequence given the recurrence relations, ensuring a consistent repeat every 3 steps.