Is there a real root?

Let $f(x) = nx^4+4x+3$ . Let us check the number of inflection points $f(x)$ has. $\dfrac {f(x)}{dx} = 4nx^3 + 4 = 0$ , $\implies x_i = - \dfrac 1{\sqrt[3]n}$ . Therefore there is only one inflection point. And since $\dfrac {d^2f(x)}{dx^2} = 12nx^2 > 0$ , $f(x_i)$ is a minimum. For $f(x)$ to have real roots,

$\begin{aligned} f(x_i) & \le 0 \\ n \left(-\frac 1{\sqrt[3] n}\right)^4 - \frac 4{\sqrt[3] n} + 3 & \le 0 \\ - \frac 3{\sqrt[3] n} + 3 & \le 0 \\ \implies \sqrt[3]n & \le 1 \\ n & \le 1 \end{aligned}$

Therefore there is only $\boxed 1$ positive integer $n$ the equation $nx^4+4x+3=0$ has a real root.

$f'(x)=4x^3+4$ , so $min{f(x)}=f(-1)=n-1$ , which is positive for $n>1$

For $n=1$ equation has root $x=-1$

If $n = 1$ , $x^4 + 4x + 3 = 0, x = -1, x = 1$ .

However, if $n \geq 2$ , there is no real root of $x$ .

Therefore, the answer is $\fbox 1$ .