Abstract: Let n be a positive integer satisfying n＞1  and s(n)=[n/2],where s(n) is the sum of the aliquot parts of n.Further let ε(n) denote the number of dirts of n.Further let ε(n) denote the number of distinct prime factors of n and p1，p2，…，pω(n) denote its prime factors with p1＜p2＜…＜pω(n).In this paper we prove that if 2｜n,then n=2;if n is odd and ω(n)≤2,then n is a power of 3;if n is odd and ω(n)≤3,then p1=3 or p1=5,p2=7 and 11≤p3≤31；if n is odd and ω(n)=4，then p1=3 or p1=5,7≤p2≤13,11≤p3≤17 and 13≤p4≤23.The above-mentioned results partly solve a problem posed by Graham.

Key words: aliquot part, sum of aliquot parts, Graham's problem