Given a positive integer $n$, there are $n$ boxes $B_1,...,B_n$. The following procedure can be used to add balls.
$$\text{(Procedure) Chosen two positive integers }n\geq i\geq j\geq 1\text{, we add one ball each to the boxes }B_k\text{ that }i\geq k\geq j.$$
For positive integers $x_1,...,x_n$ let $f(x_1,...,x_n)$ be the minimum amount of procedures to get all boxes have its amount of balls to be a multiple of 3, starting with $x_i$ balls for $B_i(i=1,...,n)$. Find the largest possible value of $f(x_1,...,x_n)$. (If $x_1,...,x_n$ are all multiples of 3, $f(x_1,...,x_n)=0$.)
To solve the problem, we need to determine the minimum number of procedures required to make the number of balls in each box a multiple of 3. We will use the properties of numbers modulo 3 to simplify the problem.

1. **Consider the numbers modulo 3:**
   Each box \( B_i \) initially contains \( x_i \) balls. We are interested in the values of \( x_i \mod 3 \). Let \( A \) be the number of boxes with \( x_i \equiv 0 \pmod{3} \), \( B \) be the number of boxes with \( x_i \equiv 1 \pmod{3} \), and \( C \) be the number of boxes with \( x_i \equiv 2 \pmod{3} \).

2. **Define the procedure:**
   The procedure allows us to add one ball to each box in a contiguous subarray \( B_j, B_{j+1}, \ldots, B_i \). This operation changes the values of \( x_j, x_{j+1}, \ldots, x_i \) by 1 modulo 3.

3. **Algorithm 1:**
   - Change all \( B \) boxes with \( x_i \equiv 1 \pmod{3} \) to \( x_i \equiv 2 \pmod{3} \). This requires \( B \) operations.
   - Change consecutive subsequences of boxes with \( x_i \equiv 2 \pmod{3} \) to \( x_i \equiv 0 \pmod{3} \). Since there are \( A \) boxes with \( x_i \equiv 0 \pmod{3} \), there can be at most \( A+1 \) such subsequences. This requires at most \( A+1 \) operations.
   - Total operations: \( A + B + 1 \).

4. **Algorithm 2:**
   - Change all \( C \) boxes with \( x_i \equiv 2 \pmod{3} \) to \( x_i \equiv 0 \pmod{3} \). This requires \( C \) operations.
   - Change consecutive subsequences of boxes with \( x_i \equiv 0 \pmod{3} \) to \( x_i \equiv 1 \pmod{3} \). Since there are \( B \) boxes with \( x_i \equiv 1 \pmod{3} \), there can be at most \( B-1 \) such subsequences. This requires at most \( B-1 \) operations.
   - Change all \( B \) boxes with \( x_i \equiv 1 \pmod{3} \) to \( x_i \equiv 0 \pmod{3} \). This requires \( B \) operations.
   - Total operations: \( C + B - 1 + B = C + 2B - 1 \).

5. **Algorithm 3:**
   - Change all \( A \) boxes with \( x_i \equiv 0 \pmod{3} \) to \( x_i \equiv 1 \pmod{3} \). This requires \( A \) operations.
   - Change consecutive subsequences of boxes with \( x_i \equiv 1 \pmod{3} \) to \( x_i \equiv 2 \pmod{3} \). Since there are \( C \) boxes with \( x_i \equiv 2 \pmod{3} \), there can be at most \( C+1 \) such subsequences. This requires at most \( C+1 \) operations.
   - Change all \( C \) boxes with \( x_i \equiv 2 \pmod{3} \) to \( x_i \equiv 0 \pmod{3} \). This requires \( C \) operations.
   - Total operations: \( A + C + 1 + C = A + 2C + 1 \).

6. **Determine the maximum number of operations:**
   We need to find the maximum value among \( A + B + 1 \), \( C + 2B - 1 \), and \( A + 2C + 1 \). Note that the sum of these values is \( 2(A + B + C) + 4 = 2n + 4 \).

7. **Use the pigeonhole principle:**
   Since the sum of the three values is \( 2n + 4 \), at least one of them must be less than or equal to \( \frac{2n + 4}{3} \). Therefore, the maximum number of operations is bounded by \( \left\lceil \frac{2n + 4}{3} \right\rceil \).

8. **Calculate the largest possible value of \( f(x_1, \ldots, x_n) \):**
   - For \( n = 3m \), \( \left\lceil \frac{2(3m) + 4}{3} \right\rceil = \left\lceil 2m + \frac{4}{3} \right\rceil = 2m + 1 \).
   - For \( n = 3m + 1 \), \( \left\lceil \frac{2(3m + 1) + 4}{3} \right\rceil = \left\lceil 2m + \frac{6}{3} \right\rceil = 2m + 2 \).
   - For \( n = 3m + 2 \), \( \left\lceil \frac{2(3m + 2) + 4}{3} \right\rceil = \left\lceil 2m + \frac{8}{3} \right\rceil = 2m + 2 \).

Therefore, the largest possible value of \( f(x_1, \ldots, x_n) \) is:
- \( 2m + 1 \) for \( n = 3m \),
- \( 2m + 2 \) for \( n = 3m + 1 \),
- \( 2m + 2 \) for \( n = 3m + 2 \).

The final answer is \( \boxed{2m + 1 \text{ for } n = 3m, \text{ and } 2m + 2 \text{ for } n = 3m + 1 \text{ or } 3m + 2} \).