我需要一个可以运行的完整代码，问题是分成6个线程，计算m到n的值(以1到100000000为例)的总和。要求使用Executors框架，每个线程计算的数字量之差不超过1

以下是一个可以运行的Java代码，使用了Executors框架，将计算任务分成6个线程，每个线程计算的数字量之差不超过1，最终将结果累加得到总和。

import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.TimeUnit;

public class SumCalculator {
    private static final int THREAD_COUNT = 6;
    private static final long START_NUMBER = 1;
    private static final long END_NUMBER = 100000000;

    public static void main(String[] args) throws InterruptedException {
        ExecutorService executor = Executors.newFixedThreadPool(THREAD_COUNT);
        long batchSize = (END_NUMBER - START_NUMBER + 1) / THREAD_COUNT;
        long remainder = (END_NUMBER - START_NUMBER + 1) % THREAD_COUNT;
        long currentNumber = START_NUMBER;
        long sum = 0;

        for (int i = 0; i < THREAD_COUNT; i++) {
            long batchSizeForThread = batchSize + (i < remainder ? 1 : 0);
            executor.submit(new SumCalculatorTask(currentNumber, currentNumber + batchSizeForThread - 1));
            currentNumber += batchSizeForThread;
        }

        executor.shutdown();
        executor.awaitTermination(1, TimeUnit.MINUTES);

        sum = SumCalculatorTask.getSum();
        System.out.println("Sum of " + START_NUMBER + " to " + END_NUMBER + " is: " + sum);
    }

    private static class SumCalculatorTask implements Runnable {
        private final long startNumber;
        private final long endNumber;
        private static volatile long sum = 0;

        public SumCalculatorTask(long startNumber, long endNumber) {
            this.startNumber = startNumber;
            this.endNumber = endNumber;
        }

        @Override
        public void run() {
            long localSum = 0;
            for (long i = startNumber; i <= endNumber; i++) {
                localSum += i;
            }
            sum += localSum;
        }

        public static long getSum() {
            return sum;
        }
    }
}

代码中首先定义了线程数量和计算的起始和结束数字，然后创建了一个固定线程数的线程池，并根据线程数量和计算范围计算出每个线程需要处理的数字数量。接着通过循环创建多个线程，并将计算任务分配给它们处理。最后等待所有线程完成任务，获取每个线程的计算结果并将它们累加得到最终的总和。