Let $$\binom{n}{k}=\frac{n!}{k!(n-k)!}$$ be the binomial coefficient where $n\ge 1$, $0\le k\le n$. I am wondering if the following is true
Let $n\ge 1$ be integers and for any $0\le k\le n$, $\lambda_k\in\mathbb{N}$. Then $$\sum_{k=0}^n \binom{n}{k}(-1)^{\lambda_k}=0$$ if and only if $\lambda_k\equiv k\mod 2$ or $\lambda_k\equiv k+1\mod 2$.
In non-mathematical terms, the only way you can make a $\pm 1$-combination of binomial coefficients to be $0$ is that you arrange the $+1$ and $-1$ alternatively.
The "if" part follows directly from binomial expansion of $$0=(1-1)^n=\sum_{k=0}^n\binom{n}{k}(-1)^k.$$
Here is a somewhat less trivial counterexample than the ones I put in the comments: Let $n=14$, let the exponents be $1,-1,1,-1,-1,1,-1,-1,1,1,1,-1,1,-1,1$.
In numbers, this is $$1-14+91-364-1001+2002-3003-3432+3003+2002+1001-364+91-14+1=0$$