可以通过以下代码的思路做，或者自己写一个轮询工厂：

   @Scheduled(cron = "0 0/3 * * * ?")
    private static void print(){
        flagA=0;
        flagB=0;
        System.out.print("将A、B清0\n");
    }

//   private static ApplicationContext context = new ClassPathXmlApplicationContext("applicationContext.xml");
//   private static UserRmiService userRmiService = (UserRmiService) context.getBean("loadBalancedService");


   private static UserRmiService load(){
       if (Math.random()<0.5 && flagA==0) {
           try {
               System.out.print("准备调用:");
               userRmiService = (UserRmiService) context.getBean("userRmiServiceA");
               System.out.print("调用成功8080端口"+userRmiService+"\n");
           }
           catch (Exception e ){
               flagA = 1;
               userRmiService = (UserRmiService) context.getBean("userRmiServiceB");
               System.out.print("调用失败尝试调用:"+userRmiService+"\n");
           }
       }
       else  if (flagB==0){
           try {
               System.out.print("准备调用:");
               userRmiService = (UserRmiService) context.getBean("userRmiServiceB");
               System.out.print("调用成功8082端口"+userRmiService+"\n");
           }
           catch (Exception e){
               flagB = 1;
               userRmiService = (UserRmiService) context.getBean("userRmiServiceA");
               System.out.print("调用失败尝试调用:"+userRmiService+"\n");
           }
       }
       else if(flagA==1 && flagB ==1) {
           System.out.print("两个端口都挂了,你在搞毛线\n");
       }
       return userRmiService;
   }

使用随机数和try catch