Can all linear lambda calculi be linearity checked syntactically?
I don't want to make a statement about "all linear lambda calculi" since it's hard to make that precise, but for pure linear lambda-calculus the answer is yes.  One way to enforce linearity in pure lambda calculus is by trying to type application and abstraction using the linear implication $A \multimap B$,
$$
\frac{\Gamma \vdash t:A\multimap B\quad \Delta\vdash u:A}{\Gamma,\Delta\vdash t(u):B}
 \qquad
\frac{x:A,\Gamma \vdash t:B}{\Gamma\vdash \lambda x.t:A\multimap B}
$$
together with the linear identity and exchange rules:
$$
\frac{}{x:A\vdash x:A}
 \qquad
\frac{\Gamma,y:B,x:A,\Delta \vdash t:C}{\Gamma,x:A,y:B,\Delta \vdash t:C}
$$

But if you just want to check that a lambda term is linear, types are overkill, and it suffices to check syntactically that every free or $\lambda$-bound variable is used exactly once.

A way to make that a bit more precise is to define the set of linear lambda terms $\Lambda_1(\gamma)$ with a given list $\gamma = x_i,\dots,x_1$ of distinct free variables.  This is really a family of sets (indexed by $\gamma$) generated by the following rules (which can be seen as an abstraction of the typing rules above):
$$
\frac{}{x \in \Lambda_1(x)}
 \qquad
\frac{t \in \Lambda_1(\gamma,y,x,\delta)}{t \in \Lambda_1(\gamma,x,y,\delta)}
 \qquad
\frac{t\in \Lambda_1(\gamma)\quad u \in \Lambda_1(\delta)}{t(u) \in \Lambda_1(\gamma,\delta)}
 \qquad
\frac{t \in \Lambda_1(x,\gamma)}{\lambda x.t \in \Lambda_1(\gamma)}
$$
Then a term $t$ of pure lambda calculus with free variables $\gamma$ is linear just in case $t \in \Lambda_1(\gamma)$.  Note that these rules (like the typing rules) also have the effect of performing scope checking, and thus ruling out certain syntactic expressions which might look linear but violate scoping (e.g., an expression like "$\lambda x.x(y)(\lambda y.\lambda z.z)$", where the variable $y$ is used before it is bound).

Finally, pure linear lambda calculus has the remarkable property that every term is typable (which is related to the fact that it is strongly normalizing).  So, once we know that a lambda term is linear it is always possible to go back and give it a type.  Even more remarkably, the principal type of a term uniquely identifies its $\beta$-normal form, so type inference is in fact equivalent to normalization.  Harry Mairson had a nice paper discussing some aspects of these remarkable properties in Linear lambda calculus and PTIME-completeness (JFP 14:6, 2004).