The set of $n$-qubit Pauli matrices $\{P_i\}_{i=1,\dots,4^n}$ forms a (orthogonal) basis for the vector space of complex matrices.
In particular, the set of Pauli matrices is linearly independent. Hence, the equation
$$
\sum_{i=1}^{4^n} x_i P_i = 0,
$$
does only have a trivial solution $x_i=0$. This implies that the sum over any subset cannot be the zero matrix (this would correspond to a solution where at least one $x_i =1$).
Definition: A $n$-qubit Pauli matrix $P$ is any tensor product
$$
P = \sigma_1\otimes\dots\otimes\sigma_n,
$$
where $\sigma_j \in \{ \mathrm I, X, Y, Z\}$.
Orthogonality: Note that $X,Y,Z$ are traceless. Moreover, we have $\sigma^2 = I$ for any $\sigma\in \{ \mathrm I, X, Y, Z\}$ and the product $\sigma\sigma'$ is, up to a phase, either $X$, $Y$, or $Z$ if $\sigma,\sigma'\in \{ \mathrm I, X, Y, Z\}$ and $\sigma\neq\sigma'$. Hence
$$
\mathrm{tr}( \sigma\sigma' ) = \begin{cases} 2 & \text{if } \sigma=\sigma' \\ 0 & \text{else} \end{cases}.
$$
Hence, the 1-qubit Pauli matrices are orthogonal w.r.t. the Hilbert-Schmidt (trace) inner product. This readily generalizes to the $n$-qubit cases which shows that the $4^n$ Pauli matrices form an orthogonal basis for the $4^n$-dimensional vector space of complex matrices $\mathbb{C}^{2^n \times 2^n}$.