What is $V^\mu$ if $\nabla_{\mu} V^{\mu}$=scalar? Suppose there is a quantity written as $\sum\limits_\mu \nabla_\mu V^\mu$ which is invariant under a coordinate transformation, i.e. scalar, where $V^\mu=(V^0,V^1,V^2,V^3)$ and $\nabla_\mu$ is a covariant derivative. Must $V^\mu$ be necessarily a vector or something else?
 A: In general, the statement that $\nabla_\mu V^\mu$ transforms as a scalar does not quite fix the transformation properties of $V^\mu$.  Rather, the most general such transformation would be
$$V^\mu \mapsto V'^\mu + C^\mu,$$
where $V'^\mu = \frac{\partial x'^\mu}{\partial x^\nu} V^\nu$ is the ordinary vector transformation law, and $C^\mu$ is any quantity satisfying $\nabla_\mu C^\mu = 0$.  You can get a large class of such transformations by taking 
$$C^\mu = \epsilon^{\alpha \beta\gamma\mu}\nabla_\alpha B_{\beta\gamma},$$
for any two form $B_{\beta\gamma}$, since such a quantity will have identically vanishing divergence.  Thus, $V^\mu$ is a vector up to the "gauge ambiguity" parameterized by $B_{\beta\gamma}$, or simply another object entirely that doesn't transform covariantly. 
A: When you write $V^\mu$ you mean that $V$ is a vector. Next $\nabla_\mu V^\mu$ is called divergence of a vector. Finally answering your question, a vector field with constant zero divergence is called incompressible or solenoidal – in this case, no net flow can occur across any closed surface (according to the Gauss law). If it is a constant, but not zero, then there is a constant flow across any closed surface.
