143 is special ...

By using Bezout's theorem we can say that there exists exactly 1 positive integer $k<143$ which has the stated property . It's possible to find the number by using Bezout's theorem , as $gcd(67,143)=1$

Thus 1 can be expressed as $143x +67y$ for some integers $x$ and $y$ .

$143 = 2\times 67 + 9$

$67 = 7 \times 9 +4$

$9 = 2\times 4 +1$

Hence $1 = 9 - 4\times 2$

$\therefore 1 = (143 - 2\times 67) - (67- 7\times 9) 2$

$\therefore 1 = (143 - 2\times 67) - [2\times 67- 14(143 - 2\times 67) ]$

$\therefore 1 = 143 - 2\times 67 +14\times 143 -30 \times 67$

$\therefore 1 = 143 \times 15 - 67\times 30$

$\therefore 1 = 143 \times 15 + 67\times (-30)$

This tells us that $67\times (-30) \equiv 1 \pmod{143}$ hence $67\times (143-30) \equiv 1 \pmod{143}$

So the smallest positive integer for which given statement hold is 143-30 i.e. $111$

The other numbers will be of the form $111 +143k$ as they also will be congruent to 1 mod 143 .

The multiple of 143 next to 10000 is $143 \times 70 = 10010$ so the biggest of our required numbers is $10010-111= 111 + 69 \times 143$ .

Hence all the numbers we require are $111,111+143 , 111+2\times 143 ,\dots, 111+ 69\times 143$

Thus there are $\boxed{70}$ such numbers $k$ ............. $(k<10000)$

I have another approach...

Since k is the inverse of 67, it is unique in the interval [1, 143]. Since 143 = 11 x 13, 67k is congruent to 1 modulo11 and modulo 13. By property of LCD congruence, this holds true. Using Chinese Remainder Theorem, we found out that k = 111 + 143n so k is congruent to 111 modulo 143. From this, we solve the inequality 111 + 143(n-1) < 10000. And n <= 70 < 70 + (22/143). Hence, the answer is 70.