Solution:

Let the crooks have stolen $a$ coins worth 11 Kulotnik and $b$ coins worth 12 Kulotnik. Here $a$ and $b$ are two non-negative integers with
$$11 a + 12 b = 5940 = 11 \cdot 540 = 12 \cdot 495$$
We now assume $b > 0$ and try to find a fair division of the loot among the crooks.
By (1), $11 a = 12 \cdot (495 - b)$, i.e. $11 a$ is divisible by 12. Since 11 and 12 are coprime, it follows that $a$ is also a multiple of 12, and therefore the crooks can distribute the stolen 11-Kulotnik coins into $\frac{a}{12}$ small bags, each of which contains $12 \cdot 11 = 132$ Kulotnik. Similarly, (1) implies the equation $12 b = 11 \cdot (540 - a)$, and hence $12 b$ is divisible by 11. As before, this shows that $b$ is also divisible by 11. Since we assumed $b > 0$, each of the crooks can take a coin worth 12 Kulotnik, and the number $b - 11$ of the remaining 12-Kulotnik coins is still divisible by 11, so that these can be distributed into $\frac{b}{11} - 1$ small bags, which likewise each contain $11 \cdot 12 = 132$ Kulotnik.
Because of
$$5940 - 11 \cdot 12 = 5808 = 132 \cdot 44$$
the as yet undistributed remainder of the loot is now contained in $44 = 11 \cdot 4$ small bags of 132 Kulotnik each. If now each of the 11 crooks takes 4 of these small bags, they have altogether divided the loot fairly. But since their leader was able to prove that this is not possible at all, our assumption $b > 0$ must have been false: in other words, there was indeed no coin worth 12 Kulotnik in the safe.