The system is given by
$u_i'=-\lambda (u_i-u_{i-1}) - \lambda (u_i - u_{i+1})$
The inital conditions are given by $u_1(0)=u_1^0 \geq 0,..., u_n(0)=u_N^0\geq 0$.
We need to prove that
$\lim_{t\rightarrow +\infty}u_i(t) = \frac{1}{N}\sum_{i=1}^{N}u_i(0), \quad i=0,...,N-1$
As written the system is not closed: $u_1'$ and $u_N'$ cannot be specified this way since there is no $u_0$ or $u_{N+1}$ to use to define the second derivative. However, one can examine $\sum_{i=2}^{N-1} u_i'=\lambda \sum_{i=1}^{N-2} u_i - 2 \lambda \sum_{i=2}^{N-1} u_i + \lambda \sum_{i=3}^N u_i=\lambda u_1-\lambda u_2 + \lambda u_N - \lambda u_{N-1}$. To get the desired limiting behavior, one must then have that $u_1'+\lambda u_1 - \lambda u_2 + u_N' + \lambda u_N - \lambda u_{N-1}$ integrates out to zero in time. The standard way to achieve that is the Neumann boundary condition $u_1'=-\lambda u_1 + \lambda u_2,u_N'=-\lambda u_N + \lambda u_{N-1}$.
Once you have specified that boundary condition, one can form the associated system matrix and apply the Gerschgorin circle theorem and the symmetry to conclude that the eigenvalues are real and nonpositive. Thus the only thing that matters for the qualitative long term behavior is the eigenvector(s) with eigenvalue zero. Determine those (it will turn out that there is just one). Then exploit the symmetry to determine the coefficient of that eigenvector in the eigenbasis expansion of the initial condition (without having to compute all the other coefficients in the eigenbasis expansion). Then you're basically done.