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Your system of equations can be rewritten as follows: For integers $i$ and $j$ in $[0,n-2]$, \begin{equation} \begin{aligned} (2n r+4)f(i,j) =& f(i-1,j) + f(i,j-1) + f(i+1, j) + f(i, j+1) \\ &\text{if}\quad 0<i<i+j<n-3, \end{aligned} \tag{1}\label{1} \end{equation} \begin{equation} \begin{aligned} (2n r+3)f(i,0) &= f(i-1,0) + f(i,1) + f(i+1,0) \\ &\text{if}\quad 0<i<n-3, \end{aligned} \tag{2}\label{2} \end{equation} \begin{equation} \begin{aligned} (2n r+3)f(0,j) &= f(0,j-1) + f(1,j) + f(0,j+1) \\ &\text{if}\quad 0<j<n-3, \end{aligned} \tag{3}\label{3} \end{equation} \begin{equation} \begin{aligned} (2n r+2)f(0,0) &= f(1, 0) + f(0,1), \end{aligned} \tag{4}\label{4} \end{equation} \begin{equation} \begin{aligned} f(i,n-2-i) = 1 \quad\text{if}\quad 0\le i\le n-2, \end{aligned} \tag{5}\label{5} \end{equation} where $n:=N$ and $r:=(1-\lambda)/\lambda\in(0,\infty)$. Note that, if $n=5$, then condition \eqref{1} is vacuous.

Already for $n=6$, there are infinitely many solutions of the linear system \eqref{1}--\eqref{5}. These solutions are given by the following: \begin{equation} \begin{aligned} f(0,4)&=1\\ f(1,0)&=(12 r+2) f(0,0)-f(0,1)\\ f(1,1)&=(12 r+3) f(0,1)-f(0,0)-f(0,2)\\ f(1,2)&=(12 r+3) f(0,2)-f(0,1)-f(0,3)\\ f(1,3)&=1\\ f(2,0)&=(12 r+3) f(1,0)-f(0,0)-f(1,1)\\ f(2,1)&=(12 r+4) f(1,1)-f(0,1)-f(1,0)-f(1,2)\\ f(2,2)&=1\\ f(3,0)&=(12 r+3)f(2,0)-f(1,0)-f(2,1)\\ f(3,1)&=1\\ f(4,0)&=1 \end{aligned} \tag{6}\label{6} \end{equation} We see that, of the the $15$ variables $f(i,j)$ (for $n=6$), the four variables $f(0,0),f(0,1),f(0,2),f(0,3)$ are "free", in the sense that one can give $f(0,0),f(0,1),f(0,2),f(0,3)$ any values and then compute the corresponding values of the remaining $15-4=11$ variables according to \eqref{6}.

The following is an answer to the original version of the question.

Your system of equations can be rewritten as follows: For integers $i$ and $j$ in $[0,n-2]$, \begin{equation} \begin{aligned} (2n r+4)f(i,j) =& f(i-1,j) + f(i,j-1) + f(i+1, j) + f(i, j+1) \\ &\text{if}\quad 0<i<i+j<n-3, \end{aligned} \tag{1}\label{1} \end{equation} \begin{equation} \begin{aligned} (2n r+3)f(i,0) &= f(i-1,0) + f(i,1) + f(i+1,0) \\ &\text{if}\quad 0<i<n-3, \end{aligned} \tag{2}\label{2} \end{equation} \begin{equation} \begin{aligned} (2n r+3)f(0,j) &= f(0,j-1) + f(1,j) + f(0,j+1) \\ &\text{if}\quad 0<j<n-3, \end{aligned} \tag{3}\label{3} \end{equation} \begin{equation} \begin{aligned} (2n r+2)f(0,0) &= f(1, 0) + f(0,1), \end{aligned} \tag{4}\label{4} \end{equation} \begin{equation} \begin{aligned} f(i,n-2-i) = 1 \quad\text{if}\quad 0\le i\le n-2, \end{aligned} \tag{5}\label{5} \end{equation} where $n:=N$ and $r:=(1-\lambda)/\lambda\in(0,\infty)$. Note that, if $n=5$, then condition \eqref{1} is vacuous.

Already for $n=6$, there are infinitely many solutions of the linear system \eqref{1}--\eqref{5}. These solutions are given by the following: \begin{equation} \begin{aligned} f(0,4)&=1\\ f(1,0)&=(12 r+2) f(0,0)-f(0,1)\\ f(1,1)&=(12 r+3) f(0,1)-f(0,0)-f(0,2)\\ f(1,2)&=(12 r+3) f(0,2)-f(0,1)-f(0,3)\\ f(1,3)&=1\\ f(2,0)&=(12 r+3) f(1,0)-f(0,0)-f(1,1)\\ f(2,1)&=(12 r+4) f(1,1)-f(0,1)-f(1,0)-f(1,2)\\ f(2,2)&=1\\ f(3,0)&=(12 r+3)f(2,0)-f(1,0)-f(2,1)\\ f(3,1)&=1\\ f(4,0)&=1 \end{aligned} \tag{6}\label{6} \end{equation} We see that, of the the $15$ variables $f(i,j)$ (for $n=6$), the four variables $f(0,0),f(0,1),f(0,2),f(0,3)$ are "free", in the sense that one can give $f(0,0),f(0,1),f(0,2),f(0,3)$ any values and then compute the corresponding values of the remaining $15-4=11$ variables according to \eqref{6}.

Your system of equations can be rewritten as follows: For integers $i$ and $j$ in $[0,n-2]$, \begin{equation} \begin{aligned} (2n r+4)f(i,j) =& f(i-1,j) + f(i,j-1) + f(i+1, j) + f(i, j+1) \\ &\text{if}\quad 0<i<i+j<n-3, \end{aligned} \tag{1}\label{1} \end{equation} \begin{equation} \begin{aligned} (2n r+3)f(i,0) &= f(i-1,0) + f(i,1) + f(i+1,0) \\ &\text{if}\quad 0<i<n-3, \end{aligned} \tag{2}\label{2} \end{equation} \begin{equation} \begin{aligned} (2n r+3)f(0,j) &= f(0,j-1) + f(1,j) + f(0,j+1) \\ &\text{if}\quad 0<j<n-3, \end{aligned} \tag{3}\label{3} \end{equation} \begin{equation} \begin{aligned} (2n r+2)f(0,0) &= f(1, 0) + f(0,1), \end{aligned} \tag{4}\label{4} \end{equation} \begin{equation} \begin{aligned} f(i,n-2-i) = 1 \quad\text{if}\quad 0\le i\le n-2, \end{aligned} \tag{5}\label{5} \end{equation} where $n:=N$ and $r:=(1-\lambda)/\lambda\in(0,\infty)$. Note that, if $n=5$, then condition \eqref{1} is vacuous.

Already for $n=6$, there are infinitely many solutions of the linear system \eqref{1}--\eqref{5}. These solutions are given by the following: \begin{equation} \begin{aligned} f(0,4)&=1\\ f(1,0)&=(12 r+2) f(0,0)-f(0,1)\\ f(1,1)&=(12 r+3) f(0,1)-f(0,0)-f(0,2)\\ f(1,2)&=(12 r+3) f(0,2)-f(0,1)-f(0,3)\\ f(1,3)&=1\\ f(2,0)&=(12 r+3) f(1,0)-f(0,0)-f(1,1)\\ f(2,1)&=(12 r+4) f(1,1)-f(0,1)-f(1,0)-f(1,2)\\ f(2,2)&=1\\ f(3,0)&=(12 r+3)f(2,0)-f(1,0)-f(2,1)\\ f(3,1)&=1\\ f(4,0)&=1 \end{aligned} \tag{6}\label{6} \end{equation} We see that, of the the $15$ variables $f(i,j)$ (for $n=6$), the four variables $f(0,0),f(0,1),f(0,2),f(0,3)$ are "free", in the sense that one can give $f(0,0),f(0,1),f(0,2),f(0,3)$ any values and then compute the corresponding values of the remaining $15-4=11$ variables according to \eqref{6}.

Remark: Going to bed yesterday, I realized that $<n-3$ in \eqref{1} was probably meant to be $\le n-3$. This has been been confirmed in the comments by Peter Taylor and the OP. In another comment, the OP also asked if we have the uniqueness for the system \eqref{1}--\eqref{5} with the additional symmetry condition: $f(i,j)=f(j,i)$ for all $i,j$.

However, already for $n=5$ there is no uniqueness, even with the symmetry condition. Indeed, for $n=5$ and $r=1$, one has the following two solutions of the corrected system \eqref{1}--\eqref{5}: \begin{equation} f(0,0)= \frac{1}{533},f(0,1)= f(1,0)= \frac{6}{533},f(0,2)=f(2,0)=0, \\ f(0,3)=f(3,0)= 1, f(1,1)= \frac{77}{533},f(1,2)=f(2,1)=1 \end{equation} and \begin{equation} f(0,0)= \frac{8}{533},f(0,1)=f(1,0)= \frac{48}{533},f(0,2)=f(2,0)=1,\\ f(0,3)=f(3,0)= 1, f(1,1)= \frac{83}{533},f(1,2)=f(2,1)=1. \end{equation} (Since the set of solutions of the corrected system \eqref{1}--\eqref{5} is affine, there are then infinitely many solutions of this system. In fact, the set of solutions here is the set of all points of the straight line through the above two solutions.)

Your system of equations can be rewritten as follows: For integers $i$ and $j$ in $[0,n-2]$, \begin{equation} \begin{aligned} (2n r+4)f(i,j) =& f(i-1,j) + f(i,j-1) + f(i+1, j) + f(i, j+1) \\ &\text{if}\quad 0<i<i+j<n-3, \end{aligned} \tag{1}\label{1} \end{equation} \begin{equation} \begin{aligned} (2n r+3)f(i,0) &= f(i-1,0) + f(i,1) + f(i+1,0) \\ &\text{if}\quad 0<i<n-3, \end{aligned} \tag{2}\label{2} \end{equation} \begin{equation} \begin{aligned} (2n r+3)f(0,j) &= f(0,j-1) + f(1,j) + f(0,j+1) \\ &\text{if}\quad 0<j<n-3, \end{aligned} \tag{3}\label{3} \end{equation} \begin{equation} \begin{aligned} (2n r+2)f(0,0) &= f(1, 0) + f(0,1), \end{aligned} \tag{4}\label{4} \end{equation} \begin{equation} \begin{aligned} f(i,n-2-i) = 1 \quad\text{if}\quad 0\le i\le n-2, \end{aligned} \tag{5}\label{5} \end{equation} where $n:=N$ and $r:=(1-\lambda)/\lambda\in(0,\infty)$. Note that, if $n=5$, then condition \eqref{1} is vacuous.

Already for $n=6$, there are infinitely many solutions of the linear system \eqref{1}--\eqref{5}. These solutions are given by the following: \begin{equation} \begin{aligned} f(0,4)&=1\\ f(1,0)&=(12 r+2) f(0,0)-f(0,1)\\ f(1,1)&=(12 r+3) f(0,1)-f(0,0)-f(0,2)\\ f(1,2)&=(12 r+3) f(0,2)-f(0,1)-f(0,3)\\ f(1,3)&=1\\ f(2,0)&=(12 r+3) f(1,0)-f(0,0)-f(1,1)\\ f(2,1)&=(12 r+4) f(1,1)-f(0,1)-f(1,0)-f(1,2)\\ f(2,2)&=1\\ f(3,0)&=(12 r+3)f(2,0)-f(1,0)-f(2,1)\\ f(3,1)&=1\\ f(4,0)&=1 \end{aligned} \tag{6}\label{6} \end{equation} We see that, of the the $15$ variables $f(i,j)$ (for $n=6$), the four variables $f(0,0),f(0,1),f(0,2),f(0,3)$ are "free", in the sense that one can give $f(0,0),f(0,1),f(0,2),f(0,3)$ any values and then compute the corresponding values of the remaining $15-4=11$ variables according to \eqref{6}.

Your system of equations can be rewritten as follows: For integers $i$ and $j$ in $[0,n-2]$, \begin{equation} \begin{aligned} (2n r+4)f(i,j) =& f(i-1,j) + f(i,j-1) + f(i+1, j) + f(i, j+1) \\ &\text{if}\quad 0<i<i+j<n-3, \end{aligned} \tag{1}\label{1} \end{equation} \begin{equation} \begin{aligned} (2n r+3)f(i,0) &= f(i-1,0) + f(i,1) + f(i+1,0) \\ &\text{if}\quad 0<i<n-3, \end{aligned} \tag{2}\label{2} \end{equation} \begin{equation} \begin{aligned} (2n r+3)f(0,j) &= f(0,j-1) + f(1,j) + f(0,j+1) \\ &\text{if}\quad 0<j<n-3, \end{aligned} \tag{3}\label{3} \end{equation} \begin{equation} \begin{aligned} (2n r+2)f(0,0) &= f(1, 0) + f(0,1), \end{aligned} \tag{4}\label{4} \end{equation} \begin{equation} \begin{aligned} f(i,n-2-i) = 1 \quad\text{if}\quad 0\le i\le n-2, \end{aligned} \tag{5}\label{5} \end{equation} where $n:=N$ and $r:=(1-\lambda)/\lambda\in(0,\infty)$. Note that, if $n=5$, then condition \eqref{1} is vacuous.

Already for $n=6$, there are infinitely many solutions of the linear system \eqref{1}--\eqref{5}. These solutions are given by the following: \begin{equation} \begin{aligned} f(0,4)&=1\\ f(1,0)&=(12 r+2) f(0,0)-f(0,1)\\ f(1,1)&=(12 r+3) f(0,1)-f(0,0)-f(0,2)\\ f(1,2)&=(12 r+3) f(0,2)-f(0,1)-f(0,3)\\ f(1,3)&=1\\ f(2,0)&=(12 r+3) f(1,0)-f(0,0)-f(1,1)\\ f(2,1)&=(12 r+4) f(1,1)-f(0,1)-f(1,0)-f(1,2)\\ f(2,2)&=1\\ f(3,0)&=(12 r+3)f(2,0)-f(1,0)-f(2,1)\\ f(3,1)&=1\\ f(4,0)&=1 \end{aligned} \tag{6}\label{6} \end{equation} We see that, of the the $15$ variables $f(i,j)$ (for $n=6$), the four variables $f(0,0),f(0,1),f(0,2),f(0,3)$ are "free", in the sense that one can give $f(0,0),f(0,1),f(0,2),f(0,3)$ any values and then compute the corresponding values of the remaining $15-4=11$ variables according to \eqref{6}.

Remark: Going to bed yesterday, I realized that $<n-3$ in \eqref{1} was probably meant to be $\le n-3$. This has been been confirmed in the comments by Peter Taylor and the OP. In another comment, the OP also asked if we have the uniqueness for the system \eqref{1}--\eqref{5} with the additional symmetry condition: $f(i,j)=f(j,i)$ for all $i,j$.

However, already for $n=5$ there is no uniqueness, even with the symmetry condition. Indeed, for $n=5$ and $r=1$, one has the following two solutions of the corrected system \eqref{1}--\eqref{5}: \begin{equation} f(0,0)= \frac{1}{533},f(0,1)= f(1,0)= \frac{6}{533},f(0,2)=f(2,0)=0, \\ f(0,3)=f(3,0)= 1, f(1,1)= \frac{77}{533},f(1,2)=f(2,1)=1 \end{equation} and \begin{equation} f(0,0)= \frac{8}{533},f(0,1)=f(1,0)= \frac{48}{533},f(0,2)=f(2,0)=1,\\ f(0,3)=f(3,0)= 1, f(1,1)= \frac{83}{533},f(1,2)=f(2,1)=1. \end{equation} (Since the set of solutions of the corrected system \eqref{1}--\eqref{5} is convex, there are in fact inifinitely many solutions of this system.)

Your system of equations can be rewritten as follows: For integers $i$ and $j$ in $[0,n-2]$, \begin{equation} \begin{aligned} (2n r+4)f(i,j) =& f(i-1,j) + f(i,j-1) + f(i+1, j) + f(i, j+1) \\ &\text{if}\quad 0<i<i+j<n-3, \end{aligned} \tag{1}\label{1} \end{equation} \begin{equation} \begin{aligned} (2n r+3)f(i,0) &= f(i-1,0) + f(i,1) + f(i+1,0) \\ &\text{if}\quad 0<i<n-3, \end{aligned} \tag{2}\label{2} \end{equation} \begin{equation} \begin{aligned} (2n r+3)f(0,j) &= f(0,j-1) + f(1,j) + f(0,j+1) \\ &\text{if}\quad 0<j<n-3, \end{aligned} \tag{3}\label{3} \end{equation} \begin{equation} \begin{aligned} (2n r+2)f(0,0) &= f(1, 0) + f(0,1), \end{aligned} \tag{4}\label{4} \end{equation} \begin{equation} \begin{aligned} f(i,n-2-i) = 1 \quad\text{if}\quad 0\le i\le n-2, \end{aligned} \tag{5}\label{5} \end{equation} where $n:=N$ and $r:=(1-\lambda)/\lambda\in(0,\infty)$. Note that, if $n=5$, then condition \eqref{1} is vacuous.

Already for $n=6$, there are infinitely many solutions of the linear system \eqref{1}--\eqref{5}. These solutions are given by the following: \begin{equation} \begin{aligned} f(0,4)&=1\\ f(1,0)&=(12 r+2) f(0,0)-f(0,1)\\ f(1,1)&=(12 r+3) f(0,1)-f(0,0)-f(0,2)\\ f(1,2)&=(12 r+3) f(0,2)-f(0,1)-f(0,3)\\ f(1,3)&=1\\ f(2,0)&=(12 r+3) f(1,0)-f(0,0)-f(1,1)\\ f(2,1)&=(12 r+4) f(1,1)-f(0,1)-f(1,0)-f(1,2)\\ f(2,2)&=1\\ f(3,0)&=(12 r+3)f(2,0)-f(1,0)-f(2,1)\\ f(3,1)&=1\\ f(4,0)&=1 \end{aligned} \tag{6}\label{6} \end{equation} We see that, of the the $15$ variables $f(i,j)$ (for $n=6$), the four variables $f(0,0),f(0,1),f(0,2),f(0,3)$ are "free", in the sense that one can give $f(0,0),f(0,1),f(0,2),f(0,3)$ any values and then compute the corresponding values of the remaining $15-4=11$ variables according to \eqref{6}.

Remark: Going to bed yesterday, I realized that $<n-3$ in \eqref{1} was probably meant to be $\le n-3$. This has been been confirmed in the comments by Peter Taylor and the OP. In another comment, the OP also asked if we have the uniqueness for the system \eqref{1}--\eqref{5} with the additional symmetry condition: $f(i,j)=f(j,i)$ for all $i,j$.

However, already for $n=5$ there is no uniqueness, even with the symmetry condition. Indeed, for $n=5$ and $r=1$, one has the following two solutions of the corrected system \eqref{1}--\eqref{5}: \begin{equation} f(0,0)= \frac{1}{533},f(0,1)= f(1,0)= \frac{6}{533},f(0,2)=f(2,0)=0, \\ f(0,3)=f(3,0)= 1, f(1,1)= \frac{77}{533},f(1,2)=f(2,1)=1 \end{equation} and \begin{equation} f(0,0)= \frac{8}{533},f(0,1)=f(1,0)= \frac{48}{533},f(0,2)=f(2,0)=1,\\ f(0,3)=f(3,0)= 1, f(1,1)= \frac{83}{533},f(1,2)=f(2,1)=1. \end{equation} (Since the set of solutions of the corrected system \eqref{1}--\eqref{5} is affine, there are then infinitely many solutions of this system. In fact, the set of solutions here is the set of all points of the straight line through the above two solutions.)