$N\equiv 3\pmod 7,\ N\equiv 5\pmod{11},\ N\equiv 0\pmod{13}.$

Since $7,11,13$ are pairwise coprime, by the Chinese Remainder Theorem, $N$ is uniquely determined modulo $7\cdot 11\cdot 13 = 1001.$

Finding $N$: Write $N=13k$. Then

$$\begin{align*} 13k \equiv 3\pmod 7 &\Rightarrow 6k \equiv 3 \Rightarrow -k \equiv 3 \Rightarrow k \equiv 4\pmod 7, 13k \equiv 5\pmod{11} &\Rightarrow 2k \equiv 5 \Rightarrow k \equiv 5\cdot 6 = 30 \equiv 8 \pmod{11}. \end{align*}$$

($2^{-1}\equiv 6 \pmod{11}$ since $2\cdot 6 = 12\equiv 1$.)

Combine: $k=7m+4$, then $7m+4\equiv 8\pmod{11} \Rightarrow 7m\equiv 4 \Rightarrow m\equiv 4\cdot 8=32\equiv 10\pmod{11}$ (since $7\cdot 8 = 56\equiv 1\pmod{11}$).

So $m = 11j+10$, $k = 77j+74$. Smallest positive $k$ is $74$, giving $N = 13\cdot 74 = 962.$

Verification: $962 = 7\cdot 137 + 3$ \checkmark, $962 = 11\cdot 87 + 5$ \checkmark, $962/13 = 74$ \checkmark.

The next solution is $N + 1001 = 1963$. So $M = 1963$ and $M-N = 1001.$

Why this is CAT-level: The question looks like a standard CRT problem, but the third condition (``divisible by $13$'') is the moduli twist -- it must be expressed as $N\equiv 0\pmod{13}$. Two layers of modular inversion ($2^{-1}\pmod{11}$, $7^{-1}\pmod{11}$) are required. The clean route exploits the fact that $M-N$ is simply $\mathrm{lcm}(7,11,13)=1001$, sidestepping the entire computation of $N$ itself. Students who don't see this shortcut compute $N$ first and then add the LCM -- which works but is slower.

Answer: $M - N = 1001.$