Is is possible to find two different sets of numbers $\{ a_1, a_2, \dots, a_N\}$ and $\{ b_1, b_2, \dots, b_N\}$ with $a_i,b_i\in[-1,1]$ such that $\sum a_i = \sum b_i$ and $\sum a_i^2 = \sum b_i^2$ are both true at the same time?
EDIT: Also, no number in each set may be $0$, i.e. $a_i,b_i \neq 0$. John Omielan has posted an example for $N=3$, before I edited the question.
For $N=1$, it is obvious that this is impossible, because $a_1 = b_1$ (sum over elements in the set) and $a_1 \neq b_1$ (the sets must be different) cannot be true at the same time. Is there a minimum number of $N$ for which it is possible to find such sets, or is it never possible?
As you stated, it's not possible for $N = 1$. It's also not possible for $N = 2$, where I assume the order doesn't count. To see this, assume there is a solution to get
$$a_1 + a_2 = b_1 + b_2 = c \tag{1}\label{eq1A}$$
$$a_1^2 + a_2^2 = b_1^2 + b_2^2 = d \tag{2}\label{eq2A}$$
Squaring the first $2$ sides of \eqref{eq1A} and subtracting \eqref{eq2A} gives
$$2a_1 a_2 = 2b_1 b_2 \implies a_1 a_2 = b_1 b_2 = e \tag{3}\label{eq3A}$$
Using Vieta's formulas, or just simply expanding a quadratic, i.e., $(x - r_1)(x - r_2) = x^2 - (r_1 + r_2)x + r_1 r_2$, we have that $a_1$ and $a_2$, as well as $b_1$ and $b_2$, are the roots of
$$x^2 - cx + e = 0 \tag{4}\label{eq4A}$$
Since \eqref{eq4A} has only $2$ roots (e.g., by seeing it's a parabola or, more formally, using the Fundamental theorem of algebra), this means $a_1$ and $a_2$ must be the same, up to order, as $b_1$ and $b_2$.
However, for $N = 3$, we have
$$\frac{1}{2} - \frac{1}{2} + 0 = \frac{1}{\sqrt{3}} - \frac{1}{2\sqrt{3}} - \frac{1}{2\sqrt{3}} = 0 \tag{5}\label{eq5A}$$
$$\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + 0 = \left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{2\sqrt{3}}\right)^2 + \left(\frac{1}{2\sqrt{3}}\right)^2 = \frac{1}{2} \tag{6}\label{eq6A}$$
For any $N \gt 3$, as stated in Kavi Rama Murthy's question comment, examples can be constructed by just adding zeros.
Update: Since the question now states $0$ are not allowed as values, then as dm63's comment suggests, we can just simply use values which sum to $0$ and with $b_i = -a_i$. For example, with $N = 3$, we could use
$$\frac{1}{2} - \frac{1}{3} - \frac{1}{6} = -\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 0 \tag{7}\label{eq7A}$$
$$\left(\frac{1}{2}\right)^2 + \left(-\frac{1}{3}\right)^2 + \left(-\frac{1}{6}\right)^2 = \left(-\frac{1}{2}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{6}\right)^2 = \frac{7}{18} \tag{8}\label{eq8A}$$
To not have the values adding to $0$, have the $a_i$ add up to a value relatively close to $0$, with $b_1$ being equal to this (e.g., have $|b_1| \lt |a_i|$ for $1 \le i \le 3$). Also, have $b_2 \gt 0$ and $b_3 = -b_2$. Then, using that the sums of squares are equal, solving for $b_2$ gives
$$b_2 = \sqrt{\frac{a_1^2 + a_2^2 + a_3^2 - b_1^2}{2}} \tag{9}\label{eq9A}$$
For example, we have
$$-\frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{1}{12} + \frac{\sqrt{69}}{12\sqrt{2}} - \frac{\sqrt{69}}{12\sqrt{2}} = \frac{1}{12} \tag{10}\label{eq10A}$$
$$\left(-\frac{1}{2}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{4}\right)^2 = \left(\frac{1}{12}\right)^2 + \left(\frac{\sqrt{69}}{12\sqrt{2}}\right)^2 + \left(-\frac{\sqrt{69}}{12\sqrt{2}}\right)^2 = \frac{35}{72} \tag{11}\label{eq11A}$$
You can easily extend this to $N \gt 3$. Also, you can also ensure that all $|a_i|$ and $|b_i|$ are unique, but the algebra becomes more complicated and messy.