The thieves may have looted $a$ 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 distribution of the loot among the thieves.
From (1), we have $11 a = 12 \cdot (495 - b)$, meaning $11 a$ is divisible by 12. Since 11 and 12 are coprime, $a$ must also be a multiple of 12. Therefore, the thieves can distribute the looted 11-Kulotnik coins into $\frac{a}{12}$ bags, each containing $12 \cdot 11 = 132$ Kulotnik. Similarly, (1) implies the equation $12 b = 11 \cdot (540 - a)$, so $12 b$ is divisible by 11. As before, this shows that $b$ is also divisible by 11. Since we assumed $b > 0$, each of the thieves can take one coin worth 12 Kulotnik, and the number $b - 11$ of the remaining 12-Kulotnik coins is still divisible by 11, allowing them to be distributed into $\frac{b}{11} - 1$ bags, each containing $11 \cdot 12 = 132$ Kulotnik.
Since

$$5940 - 11 \cdot 12 = 5808 = 132 \cdot 44$$

the undistributed remainder of the loot is now in $44 = 11 \cdot 4$ bags, each containing 132 Kulotnik. If each of the 11 thieves takes 4 of these bags, they will have fairly divided the loot. However, since their leader proved that this is impossible, our assumption $b > 0$ must be false: In other words, there was indeed no coin worth 12 Kulotnik in the safe.

Remark. Some participants came up with very creative nicknames for the thieves: bank robbers, pirates, gangsters, bandits, criminals, offenders, thieves.