$ f(x)$ is a given polynomial whose degree at least 2. Define the following polynomial-sequence: $ g_1(x)\equal{}f(x), g_{n\plus{}1}(x)\equal{}f(g_n(x))$, for all $ n \in N$. Let $ r_n$ be the average of $ g_n(x)$'s roots. If $ r_{19}\equal{}99$, find $ r_{99}$.
1. Given a polynomial \( f(x) \) of degree at least 2, we define a sequence of polynomials \( g_n(x) \) as follows:
   \[
   g_1(x) = f(x), \quad g_{n+1}(x) = f(g_n(x)) \quad \text{for all } n \in \mathbb{N}.
   \]
   Let \( r_n \) be the average of the roots of \( g_n(x) \).

2. We need to find \( r_{99} \) given that \( r_{19} = 99 \).

3. To proceed, we use the fact that for any polynomial \( p(x) = \alpha_d x^d + \alpha_{d-1} x^{d-1} + \ldots \) of degree \( d \), the sum of its roots is given by Vieta's formulas:
   \[
   \text{Sum of roots} = -\frac{\alpha_{d-1}}{\alpha_d}.
   \]
   Therefore, the average of the roots is:
   \[
   r = -\frac{\alpha_{d-1}}{d \alpha_d}.
   \]

4. Let \( f(x) = a_m x^m + a_{m-1} x^{m-1} + \ldots + a_1 x + a_0 \). For a fixed but arbitrary \( n \), write:
   \[
   g_n(x) = c_N x^N + c_{N-1} x^{N-1} + \ldots,
   \]
   \[
   g_{n+1}(x) = d_M x^M + d_{M-1} x^{M-1} + \ldots.
   \]

5. Since \( g_{n+1}(x) = f(g_n(x)) \), we have:
   \[
   d_M x^M + d_{M-1} x^{M-1} + \ldots = a_m (c_N x^N + c_{N-1} x^{N-1} + \ldots)^m + a_{m-1} (c_N x^N + c_{N-1} x^{N-1} + \ldots)^{m-1} + \ldots.
   \]

6. The highest degree term on the right-hand side is \( x^{Nm} \), which is obtained by raising \( c_N x^N \) to the \( m \)-th power. Thus:
   \[
   M = Nm,
   \]
   \[
   d_M = a_m c_N^m.
   \]

7. To find \( d_{M-1} \), consider the term \( x^{M-1} = x^{Nm-1} \). One way to obtain this is by raising \( c_N x^N \) to the \( (m-1) \)-th power and multiplying by \( c_{N-1} x^{N-1} \), which appears in the multinomial with coefficient \( m \). Therefore:
   \[
   d_{M-1} = m a_m c_N^{m-1} c_{N-1}.
   \]

8. Now, compare the average \( r_n \) of \( g_n \)'s roots with that of \( g_{n+1} \):
   \[
   r_n = -\frac{c_{N-1}}{N c_N},
   \]
   \[
   r_{n+1} = -\frac{d_{M-1}}{M d_M} = -\frac{m a_m c_N^{m-1} c_{N-1}}{Nm a_m c_N^m} = -\frac{c_{N-1}}{N c_N}.
   \]

9. Since \( r_n = r_{n+1} \) for all \( n \), the sequence \( r_n \) is constant. Given \( r_{19} = 99 \), it follows that:
   \[
   r_{99} = 99.
   \]

The final answer is \(\boxed{99}\).