For how many positive integers $x$ does there exist a rectangular prism $P Q R S T U V W$, labelled as shown, with $P R=1867, P V=2019$, and $P T=x$ ?
(A) 1980
(B) 1982
(C) 1984
(D) 1983
(E) 1981
Suppose that $P Q=a, P S=b$ and $P U=c$.

Since $P Q R S T U V W$ is a rectangular prism, then $Q R=P S=b$ and $S T=Q V=P U=c$.

By the Pythagorean Theorem, $P R^{2}=P Q^{2}+Q R^{2}$ and so $1867^{2}=a^{2}+b^{2}$.

By the Pythagorean Theorem, $P V^{2}=P Q^{2}+Q V^{2}$ and so $2019^{2}=a^{2}+c^{2}$.

By the Pythagorean Theorem, $P T^{2}=P S^{2}+S T^{2}$ and so $x^{2}=b^{2}+c^{2}$.

Adding these three equations, we obtain

$$
\begin{aligned}
1867^{2}+2019^{2}+x^{2} & =\left(a^{2}+b^{2}\right)+\left(a^{2}+c^{2}\right)+\left(b^{2}+c^{2}\right) \\
1867^{2}+2019^{2}+x^{2} & =2 a^{2}+2 b^{2}+2 c^{2} \\
a^{2}+b^{2}+c^{2} & =\frac{1867^{2}+2019^{2}+x^{2}}{2}
\end{aligned}
$$

Since $a^{2}+b^{2}=1867^{2}$, then

$$
c^{2}=\left(a^{2}+b^{2}+c^{2}\right)-\left(a^{2}+b^{2}\right)=\frac{1867^{2}+2019^{2}+x^{2}}{2}-1867^{2}=\frac{-1867^{2}+2019^{2}+x^{2}}{2}
$$

Since $a^{2}+c^{2}=2019^{2}$, then

$$
b^{2}=\left(a^{2}+b^{2}+c^{2}\right)-\left(a^{2}+c^{2}\right)=\frac{1867^{2}+2019^{2}+x^{2}}{2}-2019^{2}=\frac{1867^{2}-2019^{2}+x^{2}}{2}
$$

Since $b^{2}+c^{2}=x^{2}$, then

$$
a^{2}=\left(a^{2}+b^{2}+c^{2}\right)-\left(b^{2}+c^{2}\right)=\frac{1867^{2}+2019^{2}+x^{2}}{2}-x^{2}=\frac{1867^{2}+2019^{2}-x^{2}}{2}
$$

Since $a, b$ and $c$ are edge lengths of the prism, then $a, b, c>0$.

Since $a^{2}>0$, then $\frac{1867^{2}+2019^{2}-x^{2}}{2}>0$ and so $1867^{2}+2019^{2}-x^{2}>0$ or $x^{2}0$, then $\frac{1867^{2}-2019^{2}+x^{2}}{2}>0$ and so $1867^{2}-2019^{2}+x^{2}>0$ or $x^{2}>2019^{2}-1867^{2}$.

Since $c^{2}>0$, then $\frac{-1867^{2}+2019^{2}+x^{2}}{2}>0$ and so $-1867^{2}+2019^{2}+x^{2}>0$ which means that $x^{2}>1867^{2}-2019^{2}$.

Since the right side of this last inequality is negative and the left side is non-negative, then this inequality is always true.

Therefore, it must be true that $2019^{2}-1867^{2}<x^{2}<2019^{2}+1867^{2}$.

Since all three parts of this inequality are positive, then $\sqrt{2019^{2}-1867^{2}}<x<\sqrt{2019^{2}+1867^{2}}$. Since $\sqrt{2019^{2}-1867^{2}} \approx 768.55$ and $\sqrt{2019^{2}+1867^{2}} \approx 2749.92$ and $x$ is an integer, then $769 \leq x \leq 2749$.

The number of integers $x$ in this range is $2749-769+1=1981$.

Every such value of $x$ gives positive values for $a^{2}, b^{2}$ and $c^{2}$ and so positive values for $a, b$ and $c$, and so a rectangular prism $P Q R S T U V W$ with the correct lengths of face diagonals.

Therefore, there are 1981 such integers $x$.