'. Six, let $n$ be a positive integer, and let the set
$M=\{(x, y) \mid x, y$ are integers, $1 \leqslant x, y \leqslant n\}$.

Define a function $f$ on $M$ with the following properties:
(a) $f(x, y)$ takes values in the set of non-negative integers;
(b) When $1 \leqslant x \leqslant n$, we have $\sum_{y=1}^{n} f(x, y)=n-1$;
(c) If $f\left(x_{1}, y_{1}\right) f\left(x_{2}, y_{2}\right)>0$, then
$\left(x_{1}-x_{2}\right)\left(y_{1}-y_{2}\right) \geqslant 0$.
Determine the number of such functions $f$, denoted by $N(n)$, and find the specific value of $N(4)$.
(Supplied by Yao Jiangan)
Six, we first prove the following general lemma.
Lemma In each cell of an $m$-row $n$-column grid, fill in a non-negative integer, the number filled in the cell at the $i$-th row and $j$-th column is denoted by $a_{ij}$, $r_i (1 \leqslant i \leqslant m)$ and $s_j (1 \leqslant j \leqslant n)$ are non-negative integers, satisfying $\sum_{i=1}^{m} r_i = \sum_{j=1}^{n} s_j$. Then, there exists a unique way to fill the numbers that satisfies the following properties:
(1) $\sum_{j=1}^{n} a_{ij} = r_i (1 \leqslant i \leqslant m)$;
(2) $\sum_{i=1}^{m} a_{ij} = s_j (1 \leqslant j \leqslant n)$;
(3) If $a_{ij} a_{kl} > 0$, then $(k-i)(l-j) \geqslant 0$.
Proof of the lemma: We use induction on $m+n$: When $m=n=1$, it is clear that $a_{11} = r_1 = s_1$. The proposition holds.
Assume the proposition holds for grids with the sum of rows and columns less than $m+n$, and consider the filling method for an $m$-row $n$-column grid.

As shown in Figure 4, the numbers in block $A_1$ have the form $a_{1j} (j \geqslant 2)$, and the numbers in block $A_2$ have the form $a_{i1} (i \geqslant 2)$. Note that $(i-1)(1-j) < 0$, so by condition (3), either all numbers in $A_1$ or $A_2$ are zero.

Also, the sum of the numbers in $A_1$ plus $a_{11} = r_1$, and the sum of the numbers in $A_2$ plus $a_{11} = s_1$, so it must be that $a_{11} = \min(r_1, s_1)$. By symmetry, assume $r_1 \leqslant s_1$, then all numbers in $A_1$ are zero.

Consider the sub-grid formed by the last $n-1$ rows of this grid, with the sums of the rows being $r_2, \cdots, r_m$ and the sums of the columns being $s_1 - r_1, s_2, \cdots, s_n$, all of which are non-negative integers, and
$$
r_2 + \cdots + r_m = (s_1 - r_1) + s_2 + \cdots + s_n.
$$

Thus, applying the induction hypothesis to this sub-grid, we know that the filling method for this part is unique.

The numbers in the first row are already determined, with only $a_{11}$ being non-zero, and the inequality $(i-1)(j-1) \geqslant 0$ holds for any $a_{ij}$. Therefore, the filling method for the entire grid also satisfies condition (3), and the proposition is proved.
We now use the lemma to solve the original problem.
The set $M$ can be viewed as an $n$-row $n$-column grid, and each function $f$ on $M$ corresponds to a filling method for the grid. Now, the column sums of the numbers in the grid are given, and the sum of all numbers in the grid is $n(n-1)$. Hence, by the lemma, any selection of $n$ non-negative integers that sum to $n(n-1)$ uniquely corresponds to a function $f$. Therefore, the answer to this problem is the number of non-negative integer solutions to the system of equations
$$
b_1 + b_2 + \cdots + b_n = n(n-1)
$$

which is
$$
N(n) = \mathrm{C}_{n(n-1) + (n-1)}^{n-1} = \mathrm{C}_{n^2 - 1}^{n-1}
$$

In particular, $N(4) = \mathrm{C}_{15}^{3} = 455$.