Question:

Step 1: Let the 2019 vectors be $\vec{v_1}, \vec{v_2}, ..., \vec{v_{2019}}$. Step 2: Given condition: Sum of any 2018 vectors = scalar multiple of the remaining vector. For example: $\vec{v_1} + \vec{v_2} + ... + \vec{v_{2018}} = k_1 \vec{v_{2019}}$ for some scalar $k_1$ Step 3: This means: $\sum_{i=1}^{2018} \vec{v_i} = k_1 \vec{v_{2019}}$ Similarly: $\sum_{i \neq j} \vec{v_i} = k_j \vec{v_j}$ for each $j$ Step 4: For this to hold for every choice of 2018 vectors (and not all vectors being scalar multiples), all vectors must be collinear. Otherwise, removing different vectors would yield sums in different directions—impossible if each equals a scalar multiple of the removed vector. Step 5: Since vectors are collinear and not all scalar multiples of each other, write $\vec{v_i} = \lambda_i \vec{u}$ where $\vec{u}$ is a unit direction and $\lambda_i$ are scalars (some positive, some negative). Step 6: From $\sum_{i \neq j} \lambda_i \vec{u} = k_j \lambda_j \vec{u}$: $(\sum_{i=1}^{2019} \lambda_i - \lambda_j) = k_j \lambda_j$ Let $S = \sum_{i=1}^{2019} \lambda_i$. Then: $S - \lambda_j = k_j \lambda_j$ This gives: $S = (k_j + 1)\lambda_j$ for all $j$ Step 7: For this to be consistent for all $j$ with different $\lambda_j$ values, we need $S = 0$. Step 8: Therefore: $\sum_{i=1}^{2019} \vec{v_i} = 0$ ∴ Answer: 0 (or magnitude = 0 )