Too many pairs

As we are excluding the pairs where $x = y$ , the sum of all products will be

$\displaystyle\sum_{j=1}^{100} \sum_{k=1}^{100} j*k - \sum_{i=1}^{100} i^{2} =$

$\displaystyle(\sum_{j=1}^{100} j)(\sum_{k=1}^{100} k) - \frac{100*(100 + 1)*(200 + 1)}{6} =$

$(\frac{100*101}{2})^{2} - 338350 = 5050^{2} - 338350 = 25164150$ ,

the last $4$ digits of which are $\boxed{4150}$ .

The value of the problem can be seen as shown:

$$

$(1+2+3+......+98+99+100)*(1+2+3+......+98+99+100)$

$$

Applying the distributive property we make all the possible pairs $(x;y)$ including the cases where $x=y$ .

$$

It's necessary to subtract numbers in the form $x^2$

$$

For $N=100$ $$ $$ $$ $\Rightarrow$ $$ $$ $$ $\frac {N*(N+1)}{2}*\frac {N*(N+1)}{2}-\frac {N*(N+1)*(2N+1)}{6}=25164150$

$$

$S=\boxed{4150}$