We will look for a number with all digits equal 1 except one equal 4. Then the sum of their pairwise products is 503 (which is a prime number). We need to find a multiple of 503 of the form $\frac{10^{10000}-1}{9}+3\cdot 10^{k}$ where $k<10000.$ The smallest non-negative solution to $\frac{10^{10000}-1}{9}+3\cdot 10^{k}\equiv 0\pmod{503}$ is $k=423$. Therefore, $n=\frac{10^{10000}-1}{9}+3\cdot 10^{423}$ satisfies the required property.

Nice solution. This is a creative problem with many many answers. Here is a solution presented in kalva.demon.co.uk using $m=528$: Answer 11...1 2112 2112 ... 2112 (960 1s followed by 10 2112s) Solution Suppose we take 980 digits to be 1 and 20 digits to be 2. Then we can take 8 pairs (2,2), 4 pairs (2,1) and 488 pairs (1,1) giving a total of 528 = 16·3·11. The sum of the digits is 1020 which is divisible by 3, so n is certainly divisible by 3. We can arrange that half the 1s and half the 2s are in odd positions, which will ensure that n is divisible by 11. Finally, n will be divisible by 16 if the number formed by its last 4 digits is divisible by 16, so we take the last 4 digits to be 2112 (=16·132). So, for example, we can take n to be 11...1 2112 2112 ... 2112, where we have 960 1s followed by 10 2112s. I have another of my own with $m=625$. Let's see other values of $m$ that people can find and hopefully post here for fun

well, i have another one, with m=1024. We proceed similiary as with $m=528$ we look for a number in wich 1024 divides the number formed by the last 10 digits then we filled up to 1000 digits with 1's and 2's, in such way that we can took the 500 products in order to sum 1024. in fact (i guess), m can be: 2048, 4096,... with a limit, of course.

This works for m=704=(2^6)*11, n=111...18118449939646464 (984 1's). (2^6)|n since the last 6 digits of n are a multiple of 64 (646464). 11 clearly divides 111...1 since there is an even number of digits which are all equal. 11 divides 8118449939646464 because 8+1+4+9+3+6+6+6=1+8+4+9+9+4+4+4. QED