For $n \ge 2$ let $q_1(x) = q_2(x) = ... = q_n(x) = 0$ and $p_2(x) = p_3(x) = ... = p_n(x) = 0$, $q_2(x) = -\frac{x+1}{3}$, $q_1(x) = \frac{x-2}{3}$, $p_1(x) = \frac{1}{\sqrt{3}}(x+1)$. Then

$$\sum_{i=1}^{n} (p_i^2(x) + q_i^3(x)) = x, \quad (1)$$
and changing $x$ by $g(x)$ for all polynomials $p_i(x)$ and $q_i(x)$, any polynomial $g(x)$ can be obtained:

$$q_2(x) = -\frac{g(x)+1}{3}, \quad q_1(x) = \frac{g(x)-2}{3}, \quad p_1(x) = \frac{1}{\sqrt{3}}(g(x)+1),$$
from (1) it is easy to see that we get what is required.

Lets show that $n = 1$ does not suit. Suppose that there exist such polynomials $p(x)$ and $q(x)$ that
$$p^2(x) + q^3(x) = x. \quad (2)$$
Changing $x$ by $x^2$ in (2), we get: $(x - p(x^2))(x + p(x^2)) = q^3(x^2)$. Suppose that both polynomials $x - p(x^2)$ and $x + p(x^2)$ have common (possibly complex) root $\alpha$, then $\alpha = p(\alpha^2) = -p(\alpha^2)$, hence $\alpha = 0$ and $p(0) = p(\alpha^2) = \alpha = 0$, $q(0) = 0$. But in such case the left-hand side of (2) is divisible by $x^2$, and the right-hand side is not, which is

impossible. So $x - p(x^2)$ and $x + p(x^2)$ have no common roots. Then each of them has to be a cube of some polynomial (with, possibly, complex coefficients). I.e. $x - p(x^2) = q_1^3(x)$ and $x + p(x^2) = q_2^3(x)$. Adding up these equalities we get
$$2x = q_1^3(x) + q_2^3(x) = (q_1(x) + q_2(x))(q_1^2(x) - q_1(x)q_2(x) + q_2^2(x)),$$
and 2 cases are possible:
1) $q_1(x) + q_2(x) = c$;
2) $q_1(x) + q_2(x) = 2cx$.
For the first case, $\frac{2x}{c} = q_1^2(x) - q_1(x)q_2(x) + q_2^2(x) = c^2 - 3q_1(x)(c - q_1(x))$, which is impossible because the degree of the left-hand side is odd and of the right-hand side is even.
For the second case $\frac{1}{c} = q_1^2(x) - q_1(x)q_2(x) + q_2^2(x) = 4c^2x^2 - 3q_1(x)(2cx - q_1(x))$, thus $\deg q_1(x) = 1$, i.e. $q_1(x) = ax + b$ ($a \neq 0$). Then the equality $p(x^2) = x - q_1^3(x)$ is impossible, because the right-hand side is the polynomial of the third degree, and the left-hand side is the polynomial of even degree.