Strangely, a whole 3 months after this problem first appeared, and all within a couple of days, four solutions came in from four different parts of the world. They were all excellent solutions, so well done Mehmet of Robert College, Turkey; Bradley of Avery Coonley School, Downers Grove, USA; Ling of Tao Nan School, Singapore and Edwinof The Leventhorpe School, Sawbridgeworth, England. We have re-produced Edwin's solution in full below.

To find the solution to this problem I started by finding expressions for the amount owned by Alan, Ben and Chris ($A$, $B$ and $C$ respectively) in terms of the total number ($T$). This gave:

$A = T / 2$

$B = T / 3$

$C = T / 6$

So $A = 3B / 2$

$A = 3C$

$B = 2C$

I then found how many were returned to the table, in terms of the number they each grabbed ($a$ , $b$ and $g$ respectively):

$(a /2 + b /3 + g /6) = (3a +2b + g ) / 6$

Then $A$, $B$ and $C$ in terms of $a$ , $b$ and $g$ :

$A = a / 2 + (3a +2b + g )/ 18$

$B = 2b / 3 + (3a +2b + g )/ 18$

$C = 5g / 6 + (3a +2b + g )/ 18$

( where $(3a +2b + g ) /18$ is an equal share of the amount returned to the table )

$A = (12a + 2b + g ) /18$

$B = (3a +14b + g ) / 18$

$C = (3a +2b + 16g ) /18$

Knowing the relationships between $A$ and $B$; $A$ and $C$; and $B$ and $C$, I found the simultaneous equations:

$15a - 38b - g = 0$

$3a - 4b - 47g = 0$

$3a - 10b + 31g = 0$

These did not have a unique solution but gave:

$b = 13g$

$a = 33g$

Putting these into the expressions for $A$, $B$, $C$ and $T$ gave:

$A = 47g / 2$

$B = 47g / 3$

$C = 47g / 6$

$T = 47g$

The lowest value of $g$ that these will give a whole number of sovereigns for is $6$, in which case:

$T = 282$

$A = 141$

$B = 94$

$C = 47$

$a = 198$

$b = 78$

$g = 6$

So Alan grabbed $198$, Ben grabbed $78$ and Chris grabbed $6$.