Safety first

If a number $x$ is 7-safe, 11-safe and 13-safe, we must have the following to all be true:

$\begin{cases} x &\equiv 3, 4 \pmod{7}\\ x &\equiv 3,4,5,6,7,8 \pmod{11}\\ x &\equiv 3,4,5,6,7,8,9,10 \pmod{13}\\ \end{cases}$

Since these mods are all primes, by Chinese Remainder Theorem we have that all solutions for $x$ are in mod $7\times 11 \times 13 = 1001$ .

We have 2 choices for the value of $x \pmod{7}$ , 6 choices for the value of $x \pmod{11}$ and 8 choices for the value of $x \pmod{13}$ . Thus, there are $2 \times 6 \times 8 = 96$ total solutions for $x \pmod{1001}$ .

We now consider the number of solutions in the below set:

$\{\{1,2,\ldots,1001\},\{1002,1003,\ldots,2002\},\ldots,\{9009, 9010, \ldots, 10010\}\}$

From above, there are $96 \times 10 = 960$ solutions, but we must subtract the number of solutions in $\{10001, 10002, \ldots, 10010\}$ . Since $10010 \equiv 0 \pmod{7}$ , $10006 \equiv -4 \pmod{7 \cdot 11 \cdot 13}$ and $10007 \equiv -3 \pmod{7\cdot 11 \cdot 13}$ . We further note that no other values can satisfy this. Thus, we must subtract these 2 solutions.

Therefore, we have a total of $960-2=\boxed{958}$ positive integers less than 10 000 which are simultaneously 7-safe, 11-safe and 13-safe.