We are given an $n$-partite graph $G$. Each partition has $n$ vertices, some of which may be isolated. Let us number the vertices in some $i^{th}$ partition as $V\_{i1},V\_{i2},...,V\_{in}$. 


Now each non-isolated vertex $V\_{ij}$ has at least one neighbor in each of the remaining $n-1$ partitions s.t. for a given numbering of the vertices, the $n$ vertices (vertex $V\_{ij}$ and its $n-1$ neighbors) form a permutation on the second index. For e.g., consider a 4-partite graph. Each partition has 4 vertices. 


Using the numbering as given above, vertex $V\_{12}$ has as neighbors vertices $V\_{21},V\_{34},V\_{43}$. Vertex $V\_{12}$ can have other neighbors as well. We need to show that the graph $G$ will always contain $n$-clique. 


A stronger claim would be to say that every vertex is part of some $n$-clique.



n-partite n-clique
False as stated: take large $n$ and for each vertex $V\_{ij}$ choose some random permutation and draw the corresponding edges. Now, we have $n^n$ possible cliques to form. Let's look at the probability that $V\_{11},\dots, V\_{nn}$ is a clique. The chance that $V\_{11}$ acquires $k$ fixed neighbors in this subgraph when its edges are drawn is $\frac{(n-1-k)!}{(n-1)!}$. Multiplying by ${n-1\choose k}$, we see that the chance to get $k$ vertices is less than $\frac 1{k!}$. Thus, if $X\_1$ is the number of acquired neighbors for $V\_{11}$, then $Ee^{X\_1}\le e^e$. Thus $Ee^{\sum X\_j}\le e^{en}$ by independence. But the clique means that $\sum X\_j\ge n(n-1)/2$, so the probability that we have a given clique is at most $e^{en}e^{-\frac{n(n-1)}2}$, which is smaller than $n^{-n}$ by an order of magnitude.


I strongly suspect that you meant something else. Actually, my first impulse when seeing your post was to say "Consider the graph consisting of isolated vertices only" but then I decided that it would be too cheap. 


