Count it!

Let , two distinct numbers are $a,b$ .

According to the question $1 \leq |a-b| \leq 10$

Let, $a-b=k \implies a=b+k$ .Then $b$ will vary from $1 \to (100-k)$ .So, total numbers of $b$ is $=100-k-1+1=100-k$ .This every $b$ will create a pair with $a$ .

So, total number of pairs are $\sum_{k=1}^{10} (100-k)=100 \times 10 -(1+2+3+....+10)=1000-55=945)$ which is our required number.

The number of ways of choosing 2 distinct integers from $1$ to $100$ inclusive such that their difference is exactly $10$ :

$(1,11); (2,12); (3,13); (4,14) .......(90,100) = 90$ ways

The number of ways of choosing 2 distinct integers from $1$ to $100$ inclusive such that their difference is exactly $9$ :

$(1,10); (2,11); (3,12); (4,13) .......(91,100) = 91$ ways

The number of ways of choosing 2 distinct integers from $1$ to $100$ inclusive such that their difference is exactly $8$ :

$(1,9); (2,10); (3,11); (4,12) .......(92,100) = 92$ ways

The number of ways of choosing 2 distinct integers from $1$ to $100$ inclusive such that their difference is exactly $7$ :

$(1,8); (2,9); (3,10); (4,11) .......(93,100) = 93$ ways

The number of ways of choosing 2 distinct integers from $1$ to $100$ inclusive such that their difference is exactly $6$ :

$(1,7); (2,8); (3,9); (4,10) .......(94,100) = 94$ ways

The number of ways of choosing 2 distinct integers from $1$ to $100$ inclusive such that their difference is exactly $5$ :

$(1,6); (2,7); (3,8); (4,9) .......(95,100) = 95$ ways

The number of ways of choosing 2 distinct integers from $1$ to $100$ inclusive such that their difference is exactly $4$ :

$(1,5); (2,6); (3,7); (4,8) .......(96,100) = 96$ ways

The number of ways of choosing 2 distinct integers from $1$ to $100$ inclusive such that their difference is exactly $3$ :

$(1,4); (2,5); (3,6); (4,7) .......(97,100) = 97$ ways

The number of ways of choosing 2 distinct integers from $1$ to $100$ inclusive such that their difference is exactly $2$ :

$(1,3); (2,4); (3,5); (4,6) .......(98,100) = 98$ ways

The number of ways of choosing 2 distinct integers from $1$ to $100$ inclusive such that their difference is exactly $1$ :

$(1,2); (2,3); (3,4); (4,5) .......(99,100) = 99$ ways

total number of ways = $90 + 91 + 92 +93 +94 + 95 +96 +97 +98 + 99 = 945$

This is Starman walking (" waiting "?) on the sky...

Between 1 and 100 (inclusive), let's start with 1, then there are $10$ subsets with 2 elements, whose difference is at most 10 with respect to 1( from 1 to 11), ( $\space \{1,2\},...., \{1,11\}$ ).

Without counting the number 1 because we have already counted it,let's start with 2, then there are $10$ subsets with 2 elements, whose difference is at most $10$ with respect to 2 (from 2 to 12), ( $\space \{2,3\},...., \{2,12\}$ )...

Let's continue this way until the number $90$ , then there are $10$ subsets with 2 elements, whose difference is at most 10 (from 90 to 100), ( $\space \{90,91\},...., \{90,100\}$ ). Now ,we can continue with the number 91, then there are $9$ subsets with 2 elements,whose difference is at most 10, ( $\space \{91,92\}, ... , \{91,100\}$ ).

With the number 92, then there are $8$ subsets with 2 elements, whose difference is at most 10 ( $\space \{92,93\}, ... , \{92,100\}$ )...

With the number 99, then there are $1$ subset with 2 elements ( $\{99,100\}$ )

Therefore, the final answer is $90 \cdot 10 + 9 + 8 + ... + 1 = 900 + 45 = 945$