Solution:

The answer is no. If $x \geq 1$ is an integer,
$$\begin{aligned} & x(x+1)(x+2)(x+3) \\ & = [x(x+3)] \cdot [(x+1)(x+2)] \\ & = \left[x^2 + 3x\right] \cdot \left[x^2 + 3x + 2\right] \\ & = \left[\left(x^2 + 3x + 1\right) - 1\right] \cdot \left[\left(x^2 + 3x + 1\right) + 1\right] \\ & = \left(x^2 + 3x + 1\right)^2 - 1. \end{aligned}$$
Therefore the product of four consecutive positive integers is always one less than a square, and therefore cannot be a square since two positive squares cannot differ by $1$.