Given that $\begin{cases} 1981 + p = q^2 & ...(1) \\ 1981 + q = p^2 & ...(2) \end{cases}$

$\begin{aligned} (1) -(2): \quad p-q & q^2 - p^2 = (q-p)(q+p) \\ \implies p+q & = - 1 \\ q & = -(p+1) \end{aligned}$

$\begin{aligned} (1): \quad 1981 + p & = (-(p+1))^2 \\ 1981 + p & = p^2 + 2p + 1 \\ p^2 + p & = 1980 \\ p(p+1) & = 1980 & \small \blue{\text{Note that }q=-(p+1)} \\ -pq & = 1980 \\ pq & = - 1980 \\ \implies 1990+pq & = 1990 - 1980 = \boxed{10} \end{aligned}$

Not to follow approach

for p in range(-1000,10000):
    for q in range(-1000,10000):
        if (1981+p==q*q):
            if (1981+q==p*p):
                print (p,q)

Output:

-45 44
44 -45

In both cases, $pq=-1980$

$\implies 1990-1980=\boxed{10}$

$1981+p=q^2.......[1]$ $1981+q=p^2.......[2]$ $[2]-[1]\Rightarrow q-p=p^2-q^2=(p-q)(p+q)$ As $p$ and $q$ are distinct $\Rightarrow p-q\neq0$ $\Rightarrow (p+q)=\frac{q-p}{p-q}=-1\Rightarrow p+q=-1 \Rightarrow p=-(q+1)$ Putting this in $[1]$ $1981-q-1=q^2$ $\boxed{1980=(q^2+q)}$ $pq=-(q+1)q=-(q^2+q)=-1980$ $1990+pq=1990-1980=\boxed{10}$