How many?

Observe the following

• $7^a$ ends in $7$ if $a = 4k+1$ for $k \in \mathbb{N}$ .
• $7^a$ ends in $9$ if $a = 4k+2$ for $k \in \mathbb{N}$ .
• $7^a$ ends in $3$ if $a = 4k+3$ for $k \in \mathbb{N}$ .
• $7^a$ ends in $1$ if $a = 4k$ for $k \in \mathbb{N}$ .

We know that the divisibility rule for $5$ requires the number to have a $0$ or a $5$ in its unit's place. Therefore, such combinations of powers of $7$ that we can have is

• $7^{4k+1} + 7^{4k+3}$ ending with $0$ .
• $7^{4k+2} + 7^{4k}$ ending with $0$ .

Now

• For $m=4k+1$ , we get $25$ values for $m$ from $1$ to $97$ and for $n=4k+3$ , we get $25$ values for $n$ from $3$ to $99$ .
• For $m=4k+2$ , we get $25$ values for $m$ from $2$ to $98$ and for $n=4k$ , we get $25$ values for $n$ from $4$ to $100$ .

So the number of possible pairs for $m$ and $n$ in each case is $25 \cdot 25 = 625$ . And thus the total of both the cases is $625 \cdot 2 = 1250$ .

Also, since the pairs are ordered, therefore each pair $(i,j)$ in this case has another alike pair $(j,i)$ (note that all ordered pairs will have distinct members since we can never have $i = j$ ). Thus the total number of pairs comes out to be $1250 \cdot 2 = \boxed{2500}$ .