(related to Problem: Chessboard Solitaire)
Play as follows: $7—15,$ $8—16,$ $8—7,$ $2—10,$ $1—9,$ $1—2,$ $5—13,$ $3—4,$ $6—3,$ $11—1,$ $14—8,$ $6—12,$ $5—6,$ $5—11,$ $31—23,$ $32—24,$ $32—31,$ $26—18,$ $25—17,$ $25—26,$ $22—32,$ $14—22,$ $29—21,$ $14—29,$ $27—28,$ $30—27,$ $25—14,$ $30—20,$ $25—30,$ $25—5.$ The two counters left on the board are $25$ and $19$ — both belonging to the same group, as stipulated — and $19$ has never been moved from its original place.
I do not think any solution is possible in which only one counter is left on the board.
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