C :

#include<stdio.h> 

int main(){ 
	int n;
	scanf("%d",&n);
	
	int a[n],b[n],i;
	for(i=0;i < n;i++){
		scanf("%d",&a[i]);
	}
	
	//求出数各个数位的和放在数组中 
	for(i = 0;i < n;i++){
		b[i] = a[i] / 10000+a[i] / 1000 % 10 + a[i] / 100 % 10+ a[i] / 10 % 10+a[i] % 10;
	}
	
	
	//排序
	int j,k,temp; 
	for(i = 0;i < n-1;i++){
		k = i;
		for(j = i + 1;j < n;j++){
			if(b[j] < b[k]) k = j;
		} 
		
		if(k != i){
			temp = b[i];
			b[i] = b[k];
			b[k] = temp;
		}
	} 
	
	
	
	
	for(i = 0;i < n;i++){
		printf("%d ",b[i]);
	}


	return 0; 
	  
} 

C++ :

#include <iostream>
#include <iomanip>
#include <cmath>

using namespace std;

int main(){
	int n,i,j,f[30000],a[30000],t;
	cin>>n;
	for(i=0;i<n;i++){
		cin>>f[i];
		while(f[i]!=0){
			a[i]+=f[i]%10;
			f[i]/=10;
		}
	}
	for(i=0;i<n;i++){
		for(j=0;j<n-i-1;j++){
			if(a[j]>a[j+1]){
				t=a[j];
				a[j]=a[j+1];
				a[j+1]=t;
			}
		}
	}
	for(i=0;i<n;i++){
		cout<<a[i]<<" "; 
	}
}

Python :

def qiuHe(n):
   s = 0
   while n != 0:
      s = s + n % 10;
      n = n // 10;
   return s;

n=int(input())
l1=list(map(int,input().split()))
l2=[]
# for l in l1:
   # //l2.append(qiuHe(l));
for i in range(0,len(l1)):
   l1[i] = qiuHe(l1[i]);
l1.sort();

for x in l1:
   print(x,end=' ')