The following proofs assume 2 variables.
Proof of necessary condition:
If $(f_i, f_j)$ is the gradient of a function $F$, it means that:
$ \frac{\partial{F}}{\partial{x_i}} = f_i \\ \frac{\partial{F}}{\partial{x_j}} = f_j $
Now, if $F$ has continuous second partial derivatives, then according to Clairaut's theorem:
$ \frac{\partial^2{F}}{\partial{x_i}\partial{x_j}} = \frac{\partial^2{F}}{\partial{x_j}\partial{x_i}} $
Therefore:
$ \frac{\partial{f_i}}{\partial{x_j}} = \frac{\partial{f_j}}{\partial{x_i}} $
Proof of sufficient condition:
The function $F$, if it exists, has the property:
$ \frac{\partial{F}}{\partial{x_i}} = f_i $
By integrating with $x_j$ constant:
$ F = \int_{x_{i_0}}^{x_i} f_i \, dx_i + R(x_j) \tag{1} $
Now take partial derivatives of both sides with respect to $x_j$:
$ \frac{\partial{F}}{\partial{x_j}} = \frac{\partial}{\partial{x_j}}\int_{x_{i_0}}^{x_i} f_i \, dx_i + R'(x_j) = f_j $
Using differentiation under integral sign:
$ \frac{\partial{F}}{\partial{x_j}} = \int_{x_{i_0}}^{x_i} \frac{\partial{f_i}}{\partial{x_j}} \, dx_i + R'(x_j) = f_j $
Using the assumption that $\displaystyle \dfrac{\partial f_i}{\partial x_j} = \dfrac{\partial f_j}{\partial x_i}$:
$ \frac{\partial{F}}{\partial{x_j}} = \int_{x_{i_0}}^{x_i} \frac{\partial{f_j}}{\partial{x_i}} \, dx_i + R'(x_j) = f_j $
Which we can write as:
$ \left. f_j \right|_{x_{i_0}}^{x_i} + R'(x_j) = f_j $
Therefore:
$ R'(x_j) = f_j(x_{i_0}, x_j) $
And:
$ R(x_j) = \int_{x_{j_0}}^{x_j} f_j \, dx_j $
Plug in back into (1): $ F = \int_{x_{i_0}}^{x_i} f_i \, dx_i + \int_{x_{j_0}}^{x_j} f_j \, dx_j $
Therefore, we have shown that $F$ exists.