 Solution

No. For every $n\ge 2$, takein ${\mathbb{R}}^n$. These form a minimally dependent vector family: the full family is dependent because $a_n-{\sum }_{j=1}^{n-1}{a}_j=0$.

If the omitted vector is $a_n$, the remaining coordinate vectors are independent. If the omitted vector is $e_k$, a relation among the retained vectors has an $e_k$ coordinate coming only from $a_n$, so its coefficient must be zero; the coefficients of all retained coordinate vectors must then be zero. Thus every $(n-1)$-element subfamily is linearly independent, as required for a counterexample. When $n=1$, the empty subfamily is independent while the one-vector family consisting of the zero vector is not, so the answer remains negative.