Consider a circle with circumference $360$, and start from a point $P$ on the circle. Take an arc $PC_1$ that has length equal to the first coin's value. Then let $\overarc{C_1C_2}$ be as long as the value of the second coin, and so on, until $C_{241}=P$. Consider the $360$ points that have an integer distance from $P$ measured on the circumference, let them be $A_1=P$, $A_2$, ..., $A_{360}$. Now look at the regular triangles $A_iA_{i+120}A_{i+240}$ (define $A_{i+360}:=A_i$) for $i=1,2,\dots,120$. Since all $241$ points $C_j$ are amongst the points $A_i$, distributing the $241$ points between the $120$ regular triangles, we find that there are three points $A_i,A_{i+120},A_{i+240}$ that are also from $C_j$. These three points cut the circle into three equal arcs, each of which is the sum of a number of coin values. This proves that the partition into three groups of equal value is possible.